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 merry-go-round on a playground consists of a horizontal solid disk with a weight of 810 N and a radius of 1.56 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s
Answer:
Kinetic Energy of the disk = 252 J
Explanation:
weight of disk = 810 N
radius = 1.56 m
applied force = 49 N
time = 2.95 s
kinetic energy of disk = ?
first, we find the mass of the disk
mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2
mass of disk = 82.57 kg
torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m
moment of inertia I = m[tex]r^{2}[/tex] = 82.57 x [tex]1.56^{2}[/tex] = 200.9 kg-[tex]m^{2}[/tex]
recall that
Torque T = Iα
where α = angular acceleration
76.44 = 200.9α
α = 76.44/200.9 = 0.38 m/s^2
from the equation of angular motion,
ω = ω' + αt
where ω = final angular speed
ω' = initial angular speed = 0 rad/s since disk starts from rest
t = time = 2.95 s
imputing values into the equation, we have
ω = 0 + (0.38 x 2.95)
ω = 1.12 rad/s
kinetic energy of the disk = I[tex]w^{2}[/tex]
KE = 200.9 x [tex]1.12^{2}[/tex]
Kinetic Energy of the disk = 252 J
A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train
Answer:
The ratio is [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
Here we are assume the acceleration of the train is a
which makes the acceleration of each car a
From the question we are told that
Considering the second car
The force causing it s movement is mathematically represented as
[tex]F_{T2} = ma[/tex]
Considering the first car
The force causing it s movement is mathematically represented as
[tex]F = F_{T1} -F_{T2} = ma[/tex]
=> [tex]F_{T1} -ma = ma[/tex]
=> [tex]F_{T1} = 2 ma[/tex]
=> [tex]\frac{F_{T1}}{ma} = 2[/tex]
=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , determine what the minimum compression in the spring will be so that the pad does not move. Disregard the weight of the blocks and . The coefficient of friction between and the pad and between the floor and the pad is s = 0.4. Consider that the friction between the cylinder and the vertical walls is negligible
Answer: 4.08 cm.
B) Determine the lowest force required to lift the weight of 750 . The static coefficient of friction between and and between and is s= 0.25, and between and is 's = 0.5. Disregard the weight of the shims and .
Answer : 1095.4 N.
Explanation:
A) Draw free body diagrams of both blocks.
Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force N pushing up and left 10° from the vertical,
Friction force Nμ pushing down and left 10° from the horizontal,
Reaction force Fab pushing down,
and friction force Fab μ pushing left.
There are 2 forces acting on block B:
Reaction force Fab pushing up,
And elastic force kx pushing down.
(There are also horizontal forces on B, but I am ignoring them.)
Sum of forces on A in the x direction:
∑F = ma
P − N sin 10° − Nμ cos 10° − Fab μ = 0
Solve for N:
P − Fab μ = N sin 10° + Nμ cos 10°
P − Fab μ = N (sin 10° + μ cos 10°)
N = (P − Fab μ) / (sin 10° + μ cos 10°)
Sum of forces on A in the y direction:
N cos 10° − Nμ sin 10° − Fab = 0
Solve for N:
N cos 10° − Nμ sin 10° = Fab
N (cos 10° − μ sin 10°) = Fab
N = Fab / (cos 10° − μ sin 10°)
Set the expressions equal:
(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)
Cross multiply:
(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)
Distribute and solve for Fab:
P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)
P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)
Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)
Sum of forces on B in the y direction:
∑F = ma
Fab − kx = 0
kx = Fab
x = Fab / k
x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))
Plug in values and solve.
x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))
x = 0.0408 m
x = 4.08 cm
B) Draw free body diagrams of both blocks.
Force P is pushing block A to the right relative to the ground C, so friction force points to the left.
Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).
Block B moves up relative to the wall D, so friction force on B will point down.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force Fc pushing up,
Friction force Fc μ₁ pushing left,
Reaction force Fab pushing down and left 15° from the vertical,
and friction force Fab μ₂ pushing up and left 15° from the horizontal.
There are 5 forces acting on block B:
Weight force 750 n pushing down,
Normal force Fd pushing left,
Friction force Fd μ₁ pushing down,
Reaction force Fab pushing up and right 15° from the vertical,
and friction force Fab μ₂ pushing down and right 15° from the horizontal.
Sum of forces on B in the x direction:
∑F = ma
Fab μ₂ cos 15° + Fab sin 10° − Fd = 0
Fd = Fab μ₂ cos 15° + Fab sin 15°
Sum of forces on B in the y direction:
∑F = ma
-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0
Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Substitute:
(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750
Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)
Sum of forces on A in the y direction:
∑F = ma
Fc + Fab μ₂ sin 15° − Fab cos 15° = 0
Fc = Fab cos 15° − Fab μ₂ sin 15°
Sum of forces on A in the x direction:
∑F = ma
P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0
P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁
Substitute:
P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁
P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°
P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)
First, find Fab using the given values.
Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)
Fab = 1151.9 N
Now, find P.
P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)
P = 1095.4 N
What is The mass of an electron
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
A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
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]
The brakes of a car are applied to give it an acceleration of -2m/s^2. The car comes to a stop in 3s. What was its speed when the brakes were applied?
Answer:
So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s
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 tank with a constant volume of 3.72 m3 contains 22.1 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 4.5 × 104 J of energy into the gas. It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
a) What is the temperature of the gas after the energy is added?___K
b) What is the change in pressure of the gas?____Pa
c) How much work was done by the gas during this process?____J
Answer:
a) 463.29 K
b) 8065.65 Pa
c) 0 J
Explanation:
The parameters given are;
Volume of the tank, V = 3.72 m³
Number of moles of gas present in the tank, n = 22.1 moles
Temperature of the gas before heating, T₁ = 300 k
Heat added to the gas, ΔQ = 4.5 × 10⁴ J
Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole
Avogadro's number = 6.022 × 10²³ particles per mole
a) ΔQ = n × [tex]c_v[/tex] × ΔT
Where:
ΔT = T₂ - T₁
T₂ = Final temperature of the gas
Hence, by plugging in the values, we have;
4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)
[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]
T₂ = 300 + 163.29 = 463.29 K
b) The pressure of the gas is found from the relation;
P×V = n×R×T
[tex]P = \dfrac{n \times R \times T}{V}[/tex]
Where:
P = Pressure of the gas
R = Universal gas constant = 8.3145 J/(mol·K)
T = Temperature of the gas
V = Volume of the gas = 3.72 ³ (constant)
n = Number of moles of gas present = 22.1 moles (constant)
Hence the change in pressure is given by the relation;
[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]
Plugging in the values, we have;
[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]
c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;
W = f(P) × ΔV
The volume given in the question is constant
∴ ΔV = 0
Hence, W = f(P) × 0 = 0 J
No work done by the gas during the process.
A dimension is a physical nature of a quantity.
(i) give two (2) limitations of dimensional analysis..
(ii) if velocity (v), time (T) and force (F) were chosen as basic quantities, find the dimensions of mass?
Answer:
i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:
Dimesnional analysis is unable to derive relation when a physical quantity depends on more than three factors with dimensions. It is unable to derive a formula that contain exponential function, trigonometric function, and logarithmic function.ii) Given:
Velocity = v
Time = t
Force = F
Force = mass x acceleration
= mass x velocity/time
So, mass= (force x time) / velocity
[mass] = Ftv^-1
Hence, dimesnion of mass is Ftv^-1.
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]
If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocity would become constant with time. While your velocity is constant, the magnitude of the force of wind resistance is
Answer:
Mg or your weight.
Explanation:
When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.
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).
ii.
The drift velocity
(b) A 1800w toaster, a 1.3KW electric frying pan, and a 100w lamp are plugged to the same
20A, 120V circuit.
i.
What current is drawn by each device and what is the resistance of each device?
State whether this combination will blow the fuse or not.
Answer:
toaster -- 15 A, 8 Ωfry pan -- 10.83 A, 11.08 Ωlamp -- 0.83 A, 144 Ωfuse will blowExplanation:
P = VI
I = P/V = P/120
R = V/I = V/(P/V) = V^2/P = 14400/P
Toaster: I = 1800/120 = 15 . . . amps
R = 14400/1800 = 8 . . . ohms
Fry pan: I = 1300/120 = 10.833 . . . amps
R = 14400/1300 = 11.08 . . . ohms
Lamp: I = 100/120 = 0.833 . . . amps
R = 14400/100 = 144 . . . ohms
The total current exceeds 20 A, so will blow the fuse.
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.
Required:
a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.
b. Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Answer:
Explanation:
At height h , potential energy of coaster car having mass m = mgh .
The car will lose potential energy and gain kinetic energy.
height lost by car when it is at the top of loop of radius R
= h - 2R
potential energy lost = mg ( h - 2R )
kinetic energy gained = mg ( h - 2R )
kinetic energy = 0 + mg ( h - 2R )
= mg ( h - 2R )
b )
For the car to remain in contact with the track , if v be the minimum velocity
centripetal force at top = mg
m v² / R = mg
v² = gR
kinetic energy = 1/2 mv²
= 1/2 m x gR
= mgR /
If h be the minimum height that can give this kinetic energy
mg ( h - 2R ) = mgR / 2
h - 2R = R / 2
h = 2.5 R .
Superman is jogging alongside the railroad tracks on the outskirts of Metropolis at 100 km/h. He overtakes the caboose of a 500-m-long freight train traveling at 50 km/h. At that moment he begins to accelerate at 10 m/s2. How far will the train have traveled before Superman passes the locomotive?
Answer:
d = 41.91 m
Explanation:
In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:
For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:
[tex]x=x_o+v_1t[/tex] (1)
xo: initial position of the front of the train = 500m
v1: speed of the train = 50km/h
For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):
[tex]x'=v_2t+\frac{1}{2}at^2[/tex] (1)
v2: initial speed of superman = 100km/h
a: acceleration = 10m/s^2
When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:
[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]
Thus, you equal x=x'
[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]
You solve the last equation for t by using the quadratic formula:
[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]
You only use t1 = 3.02s because negative times do not have physical meaning.
Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):
[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]
x is the position of the front of the train, then, the dstance traveled by the train is:
d = 541.91m - 500m = 41.91 m
How much work must be done on a 10 kg snowboard to increase its speed from 4 m/s to 6 m/s?
Answer:
100 J
Explanation:
Work = change in energy
W = ΔKE
W = ½ mv² − ½ mv₀²
W = ½ m (v² − v₀²)
W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)
W = 100 J
The inhabitants of a small island export a cloth made from a plant that grows only on their island. A clothier from New York, believing that he can save money by "cutting out the middleman," decides to travel to the island and buy the cloth himself. Ignorant of the local custom where strangers are offered outrageous prices initially, the clothier accepts (much to everyone's surprise) the initial price of 400 tepizes/m^2. The price of this cloth in New York is 120 dollars/yard^2. If the clothing maker bought 500 m^2 of this fabric, how much money did he lose? Use 1tepiz= 0.625dollar and 0.9144m = 1yard.
Answer:
Explanation:
purchase price = 400 tepizes / m²
1 tepiz = .625 dollar
purchase price in terms of dollar = 400 x .625 dollar / m²
= 250 dollar / m²
.9144 m = 1 yard
1 m = 1.0936 yard
1m² = 1.196 yard²
price in terms of dollar / yards²
= 250 / 1.196 dollar / yard²
= 209 dollar / yard²
Price of cloth in New York = 120 dollar / yard²
loss = 209 - 120 = 89 dollar / yard²
500 m² = 500 x 1.196 yard²
= 598 yard²
net loss in purchasing 500 m² cloth
= 598 x 89
= 53222 dollar .
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
E = -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E = -4556.18 N/m
The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.
What is electric field?The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.
The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)
The horizontal distance traveled by particle is
x = ucosθt
t = x/ucosθ
The force in an electric field is F = qE...................(1)
where, q is charge , E is the strength of electric field
From, newton 2nd law of motion, Force F = ma.................(2)
Equating both the equations, we get
ma = qE
a = qE/m..................(3)
The vertical distance, y =usinθt - 1/2at²
From equation 3, we have
y = usinθt - 1/2 (qE/m) t²
if y = 0, t = 2musinθ/(qE) = x / (ucosθ)
The electric field is represented as
Also, E = 2mu²×sinθ×cosθ/(xq)
Plug the values, we get
E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)
E = -4556.18 N/m
Thus, the electric field of the particle is -4556.18 N/m.
Learn more about electric field.
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still really need help with these three questions!!
Explanation:
2. No, not always. Normal force is equal to force of gravity only when there's no acceleration in the vertical direction.
For example, when you stand in an elevator that's not moving, or moving at constant speed, then the normal force equals your weight. But when the elevator accelerates upward, the normal force increases (making you feel heavier). And when the elevator slows down, the normal force decreases (making you feel lighter).
3. Yes, it is possible for an object to be moving eastward and experience a net force westward. An example is a car applying the brakes.
4. Friction force allows you to walk. When you push against the floor, the floor's friction pushes back, as Newton's third law says.
If you try to walk on a slippery surface like ice, you won't be able to push against the ice, and the ice won't push back.
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
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?
Explanation:
It is given that,
Total mass is 70 kg
The truck exerts a constant force of 20 N.
Then the net force is given by :
F = ma
a is acceleration of rider
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]
Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :
[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]
Since, 1 m/s = 2.23 mph
4.28 m/s = 9.57 mph
So, the speed of the rider is 4.28 m/s or 9.57 mph.
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.
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
Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the spectator A is 0.55 s, and for the spectator B it is 0.45 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. The speed of sound in the air is 343 m/s.
How far are (a) spectator A and (b) spectator B from the player?
(c) How far are the spectators from each other?
Answer:
a)188.65m
b)154.35m
c)243.7m
Explanation:
Given data:
[tex]t_A=0.55s[/tex]
[tex]t_B=0.45s[/tex]
(a) The distance from the kicker to each of the 2 spectators is given by:
[tex]d_A=v \times t_A[/tex]
where,
v= speed of sound
[tex]t_A[/tex]=time taken for the sound waves to reach the ears
[tex]d_A=343\times 0.55=188.65[/tex]m
(b)[tex]d_B=v \times t_B[/tex]
where,
v= speed of sound
[tex]t_B[/tex]=time taken for the sound waves to reach the ears
[tex]d_B=343\times 0.45=154.35m[/tex]
(c)As the angle b/w slight lines from the two spectators to the player is right angle,
hypotenuse=the distance b/w 2 spectators
and, the slight lines are the other 2 lines
[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
I really need help with this question someone plz help !
Answer:D
Explanation:
Given
Same force is applied to each ball such that all have different masses
and Force is given by the product of mass and acceleration
[tex]F=m\times a[/tex]
[tex]a=\frac{F}{m}[/tex]
So acceleration of ball A
[tex]a_A=\frac{F}{0.5}=2F[/tex]
acceleration of ball B
[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]
acceleration of ball C
[tex]a_C=\frac{F}{1}=F[/tex]
acceleration of ball D
[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]
It is clear that acceleration of ball D is least.
