Answer:
The maximum speed v that the car can go without flying off the track = 15.29 m/s
Explanation:
let us first lay out the information clearly:
mass of car (m) = 410 kg
radius of race track (r) = 83.4 m
coefficient of friction (μ) = 0.286
acceleration due to gravity (g) = 9.8 m/s²
maximum speed = v m/s
For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:
F = mass × ω
where:
F = maximum centripetal force = mass × μ × g
ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]
∴ F = mass × ω
m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]
410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]
since 410 is on both sides, they will cancel out:
0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]
2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]
now, we cross-multiply the equation
2.8028 × 83.4 = [tex]v^{2}[/tex]
[tex]v^{2}[/tex] = 233.754
∴ v = √(233.754)
v = 15.29 m/s
Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 240 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
The value of F= - 830 N
Since the force is negative, it implies direction of the force applied was due south.
Explanation:
Given data:
Mass = 1000-kg
Distance, d = 240 m
Initial velocity, v1 = 20.0 m/s
Final velocity, v2 = 0 (since the car came to rest after brake was applied)
v2²= v1² + 2ad (using one of the equation of motion)
0= 20² + (2 x a x 240)
0= 400 + 480 a
a = - 400/480
a = - 0.83 m/s²
Then, imputing the value of a into
F = ma
F = 1000 kg x ( - 0.83 m/s²)
F= - 830 N
The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 950 m/s2 and the vertical or y component of its acceleration is 750 m/s2. The ball's mass is 0.35 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
F = 423.63 N
Explanation:
Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:
a = √(ax² + ay²)
where,
a = net acceleration = ?
ax = x - component of acceleration = 950 m/s²
ay = y - component of acceleration = 750 m/s²
Therefore,
a = √[(950 m/s²)² + (750 m/s²)²]
a = 1210.4 m/s²
Now, from Newton's Second Law, we know that:
F = ma
where,
m = mass of ball = 0.35 kg
F = Net force acting on ball = ?
F = (0.35 kg)(1210.4 m/s²)
F = 423.63 N
Convert from scientific notation to standard form
9.512 x 10-8
Answer:
0.00000009512
Explanation:
Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.
let's pass the number 9,512 10⁻⁸ to decimal notation
9,512 / 10⁸ = 9,512 / 100000000
0.00000009512
As we see writing this number, it is very easy to make mistakes
A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf
Answer:
0.002kg and 0.01cm
Explanation:
500 leaves has a thickness is 5cm
Means I leaf has a thickness of 5/500= 0.01cm
Similarly the mass of one leaf would be 1/500 =0.002kg
Using the equation for the distance between fringes, Δy = xλ d , complete the following. (a) Calculate the distance (in cm) between fringes for 694 nm light falling on double slits separated by 0.0850 mm, located 4.00 m from a screen. cm (b) What would be the distance between fringes (in cm) if the entire apparatus were submersed in water, whose index of refraction is 1.333? cm
Answer:
Explanation:
Distance between fringe or fringe width = xλ / d
where x is location of screen and d is slit separation
Given x = 4 m
λ = 694 nm
d = .085 x 10⁻³ m
distance between fringes
= 4 x 694 x 10⁻⁹ / .085 x 10⁻³
= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶
= 32.66 x 10⁻³ m
= 32.66 mm .
3.267 cm
b )
when submerged in water , wavelength in water becomes as follows
wavelength in water = wave length / refractive index
= 694 / 1.333 nm
= 520.63 nm
new distance between fringes
3.267 / 1.333
= 2.45 cm .
A whistle of frequency 516 Hz moves in a circle of radius 64.3 cm at an angular speed of 17.9 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle
Answer:
(a) 498.6 Hz
(b) 534.6 Hz
Explanation: Please see the attachments below
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
Find another example of separation that is used to extract a material made useful by humans. Describe the process of separation and what we use the separated component for. (4-6 sentences)
If anyone would answer this I’ll answer ur questions for return!
Please and thank you!
Answer:
Salt
Explanation:
Salt plays a crucial role in maintaining human health. It is the main source of sodium and chloride ions in the human diet. Sodium is essential for the nerve and muscle function and is involved in the regulation of fluids in the body. Sodium also plays a role in the body's control of blood pressure and volume. Salt is harvested by seawater or brine is fed into large ponds of water and is drawn out through natural evaporation which allows the salt to be subsequently harvested.
Have a good day and stay safe!
A uniform ladder stands on a rough floor and rests against a frictionless wall. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.6m, a weight of WL= 69.0N , and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the forces exerted on the ladder when the person is halfway up the ladder.
Required:
Solve of N1, N2 and f1
Answer:
The normal force N1 exerted by the floor is [tex]N_1 = 951 \ N[/tex]
The normal force N2 exerted by the wall is [tex]N_2= 616.43 \ N[/tex]
The frictional force exerted by the wall is [tex]f = N_2 = 616.43 \ N[/tex]
Explanation:
From the question we are told that
The length of the ladder is [tex]L = 4.6 \ m[/tex]
The weight of the ladder is
The distance of the ladder position on the wall from the floor is [tex]D = 3.75 \ m[/tex]
The mass of the person is [tex]m = 90 kg[/tex]
Applying Pythagoras theorem
The length of the position the ladder on the ground from the base of the wall is
[tex]A = \sqrt{L^ 2 - D^2}[/tex]
substituting values
[tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]
[tex]A = 2.66 \ m[/tex]
In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as
[tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]
[tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2= 616.43 \ N[/tex]
Now the force exerted by the floor on the ladder is mathematically represented as
[tex]N_1 = W_L + (m * g )[/tex]
substituting values
[tex]N_1 = 951 \ N[/tex]
Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so
[tex]f = N_2 = 616.43 \ N[/tex]
A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 0.9 m. What is the net force acting on a 65 kg driver who is driving at 18 m/sec and comes to rest in this distance
Answer:
11,700Newton
Explanation:
According to Newton's second law, Force = mass × acceleration
Given mass = 65kg.
Acceleration if the car can be gotten using one of the equation of motion as shown.
v² = u²+2as
v is the final velocity = 18m/s
u is the initial velocity = 0m/s
a is the acceleration
s is the distance travelled = 0.9m
On substitution;
18² = 0²+2a(0.9)
18² = 1.8a
a = 324/1.8
a = 180m/²
Net force acting on the body = 65×180
Net force acting on the body = 11,700Newton
A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS, with respect to the water. The swimmer travels a distance D in a time tOut. The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS, with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn. What is tOut in terms of vR, vS, and D, as needed?
Answer:
The time taken is [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
Explanation:
From the question we are told that
The speed of the current is [tex]v__{R}}[/tex]
The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]
The distance traveled by the swimmer is [tex]D[/tex]
The time taken to travel this distance is [tex]t_{out}[/tex]
The speed of the swimmer against direction of current is [tex]v__{s}}[/tex]
The resultant speed for downstream current is
[tex]V_{r} = v__{S}} +v__{R}}[/tex]
The time taken can be mathematically represented as
[tex]t_{out} = \frac{D}{V_{r}}[/tex]
[tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
What is the momentum of an 8kg bowling ball rolling at 2m/s
Answer:
16kg m/s
Explanation:
P=mv
8 times 2=16kg m/s
Answer:
The momentum of moving body is calculated by
p= mv
In this question m= 8kg
v= 2m/s
so p = 8*2 = 16 kg m/s.
"Mass in motion" can be used to describe momentum. Mass exists in all things. Therefore, if an object is moving, it has momentum—its mass is moving. There are two factors that determine an object's momentum level: how much and how quickly the objects are moving.
Mass and velocity are two variables that affect momentum. An object's momentum can be expressed mathematically as the product of its mass multiply by its velocity.
The equation above can be rewritten as p = m • v, where m is the mass and v is the velocity, since momentum is represented by the lower case p in physics. The equation demonstrates that an object's momentum is directly proportional to its mass and velocity.
The quantity momentum is a vector. A vector quantity is a quantity that is fully described by magnitude and direction, as was discussed in a previous unit. Information about the bowling ball's magnitude as well as its direction must be included in order to fully describe the momentum of a 5-kg ball traveling westward at 2 m/s. The ball has a momentum of 10 kg m/s.
Until information about the ball's direction is provided, the ball's momentum cannot be fully described. The direction of the ball's velocity and the direction of the momentum vector are identical. It was mentioned in a previous unit that the velocity vector moves in the same way that an object moves. The bowling ball's momentum can be fully described as 10 kg m/s westward if it is moving westward. The magnitude and direction of an object's momentum can be used to fully describe it as a vector quantity.
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A constant force applied to object A causes it to accelerate at 5 m/s2. The same force applied to object B causes an acceleration of 3 m/s2. Applied to object C, it causes an acceleration of 7 m/s2.
A. Which object has the largest mass?B. Which object has the smallest mass?C. What is the ratio of mass A to mass B?
Answer:
(A) object B has the largest mass because it has the least acceleration
(B) object C has the smallest mass because it has the largest acceleration
(C) mass A : mass B = 3 : 5
Explanation:
Given;
acceleration of object A = 5 m/s²
acceleration of object B = 3 m/s²
acceleration of object C = 7 m/s²
A constant force, F
According to Newton's second law of motion;
F = ma
m = F / a
Mass of object A:
m = F / 5
Mass of object B:
m = F / 3
Mass of object C:
m = F / 7
(A). Which object has the largest mass:
object B has the largest mass because it has the least acceleration
(B). Which object has the smallest mass:
object C has the smallest mass because it has the largest acceleration
(C). What is the ratio of mass A to mass B;
mass A = F / 5
mass B = F / 3
[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]
A. The Object B has largest mass.
B. The Object A has smallest mass.
C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]
Newton second law of motion:The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
[tex]F=ma\\\\m=\frac{F}{a}[/tex]
For constant force, mass is inversely proportional to acceleration of object.Given that, acceleration of object A is [tex]5m/s^{2}[/tex] and object B is [tex]3m/s^{2}[/tex]Thus, Object B has largest mass.Object A has smallest mass.the ratio of mass A to mass B is,[tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]
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Sara walks part way around a swimming pool. She walks 50 yards north, then
20 yards east, then 50 yards south. The magnitude of her total displacement
during this walk is
yards.
Answer:
20 Yards
Explanation:
|---20----|
| |
| 50 |50
|---D--->|
Start End
Total displacement(D) 20 yards (East).
To get up on the roof, a person (mass 69.0 kg) places a 6.40 m aluminum ladder (mass 11.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom
Answer:
N = 243.596 N ≈ 243.6 N
Explanation:
mass of person = 69 kg ( M )
mass of aluminium ladder = 11 kg ( m )
length of ladder = 6.4 m ( l )
base of ladder = 2 m from the house (d )
center of mass of ladder = 2 m from the bottom of ladder
person on ladder standing = 3 m from bottom of ladder
Calculate the magnitudes of the forces at the top and bottom of the ladder
The net torque on the ladder = o ( since it is at equilibrium )
assuming: the weight of the person( mg) acting at a distance x along the ladder. the weight of the ladder ( mg) acting halfway along the ladder and the reaction N acting on top of the ladder
X = l/2
x = 6.4 / 2 = 3.2
find angle formed by the ladder
cos ∅ = d/l
∅ = [tex]cos^{-1][/tex] 2/6.4 = [tex]cos^{-1}[/tex]0.3125 ≈ 71.79⁰
remember the net torque around is = zero
to calculate the magnitude of forces on the ladder we apply the following formula
[tex]N = \frac{mg(dcosteta)+ Mgxcosteta}{lsinteta}[/tex]
m = 11 kg, M = 69 kg, l = 6.4 , x = 3, teta( ∅ )= 71.79⁰, g = 9.8
back to equation N = [tex]\frac{11*9.8(2*cos71.79)+ 69*9.8*3* cos71.79}{6.4sin71.79}[/tex]
N = (67.375 + 633.938) / 2.879
N = 243.596 N ≈ 243.6 N
A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.200 rad/s2 , what is its angular velocity at t = 2.50 s ? (Assume the acceleration and velocity have the same direction) Express your answer in radians per second. ω = nothing rads Request Answer Part B Through what angle has the wheel turned between t = 0 and t = 2.50 s ? Express your answer in radians. Δθ = nothing rad Request Answer Provide Feedback
Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.
(A) The wheel has angular velocity at time [tex]t[/tex] according to
[tex]\omega=\omega_0+\alpha t[/tex]
so that after 2.50 s, the wheel will have attained an angular velocity of
[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]
(B) The angular displacement of the wheel is given by
[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]
After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to
[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]
In each pair, select a substance that is a better heat conductor.
1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass
Answer:
1)copper wire
Explanation:
it is the best electric conductor
A mechanic applies a force of 60N at a distance of 80 cm from the pivot on a wheel wrench. What is the size of the moment?
Answer:
48 Nm
Explanation:
Moment, or torque, is the cross product of radius and force vectors.
τ = r × F
τ = (0.80 m) (60 N)
τ = 48 Nm
can a body be in equilibrium if only one external force act on its ? explain
Answer:
Explanation:
If there is only 1 force, the body can never be in equilibrium, providing that the force is not zero (and that would hardly be a force. Zero is possible in math and it means something. It is debatable in physics).
You cannot think of a condition where something is stationary on planet earth and there are not 2 forces or more forces involved.
Think of something like a block of wood sitting on a table. It is not moving, we'll say. Gravity is holding it down, but what is pushing up on it?
The table is. There are 2 forces and they are equal in magnitude, but opposite in direction. That matters.
Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by
s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?
Answer:
a) v = -12t + 64
b) t = 5.33s
c) s = 362.66ft
d) t = 13.10s
e) v = 93.2ft/s
Explanation:
You have the following equation for the height of a stone thrown in Mars:
[tex]s(t)=-6t^2+64t+192[/tex] (1)
a) The velocity of the stone after t seconds is obtained with the derivative of s in time:
[tex]v=\frac{ds}{st}=-12t+64[/tex] (2)
The equation for the speed of the stone is v = -12t + 64
b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:
[tex]-12t+64=0\\\\t=5.33s[/tex]
The time in which the stone is at the maximum height is 5.33s
c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):
[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]
the maximum height is 362.66 ft
d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:
[tex]0=-6t^2+64t+192[/tex]
you use the quadratic formula:
[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]
You use the result with positive values because is the onlyone with physical meaning.
The time for the stone hits the ground is 13.10 s
e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:
[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]
The minus sign is because the stone's direction is downward.
The speed of the stone just when it strikes the ground is 93.2ft/s
A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.
What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.
Answer:
J = 0.800 kg m/s
Fmax = 291 N
Explanation:
During the fall, energy is conserved.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.81 m/s² × 1.45 m)
v = 5.33 m/s
Alternatively, you can use kinematics to find the velocity.
Impulse = change in momentum
J = Δp
J = mΔv
J = (0.150 kg) (5.33 m/s)
J = 0.800 kg m/s
Impulse = area under F vs t graph
J = ∫ F dt
J = ½ Fmax Δt
(0.800 kg m/s) = ½ Fmax (0.0055 ms)
Fmax = 291 N
A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?
a. 78 degree
b. 88 degree
c. 68 degree
d. 49 degree
e. the critical angle isundefined
Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
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The only force acting on a 3.2 kg canister that is moving in an xy plane has a magnitude of 6.7 N. The canister initially has a velocity of 3.3 m/s in the positive x direction, and some time later has a velocity of 6.9 m/s in the positive y direction. How much work is done on the canister by the 6.7 N force during this time
Answer:
The work done by the force is 5.76 J
Explanation:
Given;
mass of canister , m = 3.2 kg
magnitude of force, f = 6.7 N
initial velocity of the canister on x-axis, [tex]v_i[/tex]= 3.3i m/s
final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s
The work done on the canister = change in the kinetic energy of the canister
[tex]W = K.E_f - K.E_i[/tex]
where;
K.Ei is the initial kinetic energy
K.Ef is the final kinetic energy
The initial kinetic energy:
[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]
The final kinetic energy:
[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]
W = 11.04 - 5.28
W = 5.76 J
Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
b) The length of the rod:
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]
Given:
d = 1.5 mλ = 2.5 nC/m
Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.
Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]
How to calculate Electric Field?E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]
Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis
[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.
Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]
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During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.
Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
You are at a stop light in your car, stuck behind a red light. Just before the light is supposed to change, a fire engine comes zooming up towards you traveling at a horrendous 85.0 km/h. If the siren has a rated frequency 665 Hz, what frequency of the sound do you hear
Answer:
The frequency of the sound you will hear is 713.85 Hz
Explanation:
Given;
speed of your car, [tex]v_s[/tex] = 85.0 km/h
frequency of the siren, f = 665 Hz
Speed of sound in air, v = 345 m/s
The frequency of the sound you hear, can be calculated as;
[tex]f' = f(\frac{v}{v-v_s})[/tex]
Convert the speed of the car to m/s
[tex]85 \ km/h =\frac{85 \ km}{h} (\frac{1000\ m}{1 \ km})(\frac{1 \ h}{3600 \ s} ) = 23.61 \ m/s[/tex]
[tex]f' = f(\frac{v}{v-v_s} )\\\\f' = 665(\frac{345}{345-23.61} )\\\\f' = 665 (1.07346)\\\\f' = 713.85 \ Hz[/tex]
Therefore, the frequency of the sound you will hear is 713.85 Hz
A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s
Answer:
B. 9.4 s
Explanation:
In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:
Vf₁ = Vi₁ + gt₁
where,
Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)
Vi = Initial Velocity in upward motion = 36.2 m/s
g = - 9.8 m/s² (negative due to upward motion)
t₁ = Time taken in upward motion = ?
Therefore,
0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)
t₁ = (36.2 m/s)/(9.8 m/s²)
t₁ = 3.7 s
Now, using 2nd equation of motion:
h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²
where,
h₁ = distance from top of building to highest point ball reaches = ?
Therefore,
h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²
h₁ = 133.58 - 66.86 m
h₁ = 66.72 m
No, considering downward motion and using subscript 2, for it.
Using 2nd equation of motion:
h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²
where,
h₂ = height of the highest point from ground = h₁ + height of building
h₂ = 66.72 m + 90 m = 156.72 m
Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)
t₂ = Time Taken in downward motion = ?
g = 9.8 m/s²
Therefore,
156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²
t₂² = (156.72 m)/(4.9 m/s²)
t₂ = √31.98 s²
t₂ = 5.7 s
Now, the total time taken by ball to reach the ground is"
Total Time = T = t₁ + t₂
T = 3.7 s + 5.7 s
T = 9.4 s
Therefore, the correct answer is:
B. 9.4 s
How do you convert 1.3*10^6cal into joules
Answer:
5.4×10⁶J
Explanation:
1 cal = 4.184 J
1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J
An object with a mass of 1500 g (grams) accelerates 10.0 m/s2 when an
unknown force is applied to it. What is the amount of the force
Answer:
15N
Explanation:
F=ma
m=1500g = 1.5kg
a=10m/s2
1.5×10=15 N
Answer:15000gms^-2
Explanation:
F=m×a
m=1500g, a=10ms^-2
F=(1500×10)gms^-2
F=15000gms^-2
An industrial flywheel (a solid disk) of mass 10.0 kg and radius 17.3 cm is rotating at an angular speed of 22.0 rad/s. Upon being switched to a slower setting, the flywheel uniformly slows down to 13.5 rad/s after rotating through an angle of 13.8 radians. Calculate the angular acceleration of the flywheel in the process of slowing down
Answer:
Explanation:
During slowing down , initial angular velocity ω₁ = 22 rad /s
final angular velocity ω₂ = 13.5 rad /s
using the law's of motion formula for rotation
ω₂² = ω₁² + 2 αθ , α is angular acceleration and θ is angle in radian rotated during this period
13.5² = 22² - 2xα x 13.8
2xα x 13.8 = 484 - 182.25
α = 10.93 rad / s²
