Given :
A badminton shuttlecock leaves the racquet traveling 30 m/s.
It goes over the net 0.5 seconds later with a speed of 10 m/s.
To Find :
The average acceleration of the shuttlecock.
Solution :
We know, average acceleration(a) is given by :
[tex]a = \dfrac{\text{Change in speed}}{\text{Time taken }}\\\\a = \dfrac{10-30}{0.5 }\ m/s^2\\\\a = -40\ m/s^2[/tex]
Therefore, the average acceleration is -40 m/s².
A car drives 200 miles east then makes a turn and travels 50 miles north before
arriving at its destination. What is the total distance traveled by the car and what is
the displacement?
Answer:
Total distance= 250miles
Displacement= 206.15 miles
Explanation:
Step one:
given data
A car drives 200 miles east
Turns and travels 50 miles north
Step two:
The total distance travelled= 200+50= 250 miles
The displacement
S^2= 50^2+200^2
S^2= 2500+40000
S^2=42500
S=√42500
S=206.15 miles
A spring with a spring constant of 200 N/m stretches by 0.03 m. What is the potential energy of the spring?
Answer: 0.09 J
Explanation: K = 200 N/m , 1/2 X 200 N/m X (0.03 M)^2 = 0.09 J
At the start of a hockey game, the referee drops the puck between two players from opposing teams. Each player wants to push the puck in the opposite direction. For several seconds the puck does not move even though both players are pushing on it with their hockey sticks. a.) Identify the forces acting on the puck. b.) Explain why the puck does not move. c.) Describe how one of the players could get the puck to move.
Answer:
a) puck is subjected to both the forces of the hockey sticks in a horizontal direction,
b)the puck does not move since the sum of the forces is zero
c) changing the magnitude or direction of its applied force
Explanation:
a) The puck is subjected to both the forces of the hockey sticks in a horizontal direction, these forces are of equal magnitude and opposite direction since the puck is at rest.
In the direction of the y-axis (perpendicular to the ice) you have the weight of the disk and the normal to this weight that are also in equilibrium.
b) the puck does not move since the sum of the forces is zero, which implies that the forces of the hockey sticks are of equal magnitude and opposite direction.
c) the player has several ways to make the puck move
* slightly changing the angle of the club and therefore the direction of the force, in this case the disc comes out in the direction of this component
* inclined the stick slightly so that the force has a vertical component and the puck jumps in this direction
* Increasing the magnitude of the force so that the puck comes out in the opposite direction to the player
* The worst case, decreasing its force to zero and the disk comes out in its direction by the other force that had the same magnitude.
A teacher asks students to make a model of a transform plate boundary the students use blocks to represent tectonic plates and Slide the blocks past each other in the directions of the arrows as shown which event can the students best demonstrate with their models
Answer:
Hello your question is incomplete hence I will give you a general answer as regards to tectonic plates sliding past each other in a sideways direction
answer : The Transform boundary is been demonstrated by the students when sliding tectonic plates past each other in sideways directions
Explanation:
The event that can be demonstrated by the students using blocks to represent tectonic plates and sliding the clocks past each other in sideways direction is Transform Boundary of the tectonic plates
How many light years are there in one mile?
Answer:
1.70108e-13 , this is the answer hope it helps
Objects 1 and 2 attract each other with a gravitational force
of 18.0 units. If the distance separating Objects 1 and 2 is
changed to one-third the original value, then the new
gravitational force will be units.
Answer:
F' = 162 units
Explanation:
The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}\\\\[/tex]
where,
F = gravitational force = 18 units
G = Gravitational Constant
m₁ = mass of object 1
m₂ = mass of object 2
r = distance between objects
Therefore,
[tex]18 = \frac{Gm_{1}m_{2}}{r^{2}}------ eqn (1)\\\\[/tex]
Now, if we change the value of distance to one-third of original value, then:
r' = r/3
[tex]F' = \frac{Gm_{1}m_{2}}{(\frac{r}{3})^{2}}\\\\F' = (9)(\frac{Gm_{1}m_{2}}{r^{2}})[/tex]
using eqn (1):
F' = 9(18 units)
F' = 162 units
A box is sliding up an incline that makes an angle of with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.40 m/s. How far does the box travel along the incline before coming to rest
Answer:
0.231 m
Explanation:
Given that
Initial velocity, u = 1.4 m/s
Coefficient if friction, μ = 0.180
Final velocity, v = 0 m/s
Since the angle made with the inclined isn't stated. I'd be working with an angle of 15°.
The formula for the acceleration is given as
Acceleration = f.r.m.g.sinθ/m
Where f = μN = μ.m.g.cosθ
On substituting, acceleration a then is equal to
a = -(u.g.cosθ + g.sinθ)
a = -(0.18 * 9.8 * cos15 + 9.8 sin15)
a = -(1.764 * 0.9659 + 9.8 * 0.2588)
a = -(1.7038 + 2.5362)
a = -4.24 m/s²
To find the distance traveled, we use one of the equations of motions
v² = u² + 2as, with v = 0
u² = -2as
s = -u²/2a, on substituting
s = -1.4² / -2 * 4.24
s = 1.96 / 8.48
s = 0.231 m
The distance traveled by the box before coming to rest is 0.23 m.
The given parameters;
coefficient of kinetic friction, μ = 0.18initial speed of the box, u = 1.4 m/sThe normal force on the box is calculated as follows;
[tex]F_n = mgsin\ \theta[/tex]
The frictional force on the box, is calculated as follows;
[tex]F_f = \mu_k F_n\\\\[/tex]
The net force on the box is calculated as follows;
[tex]-mgsin \ \theta - \mu F_n = ma\\\\-m(gsin\theta + \mu gcos\theta) = ma\\\\- g(sin\theta + \mu cos\theta) = a\\\\-9.8(sin \ 15 \ + \ 0.18\times cos\ 15) = a\\\\-4.24 \ m/s^2 = a[/tex]
The distance traveled by the box before coming to rest is calculated as;
[tex]v^2 = u^2 + 2ad\\\\0 = u^2 + 2ad\\\\0 = (1.4)^2 + 2(-4.24)d\\\\0 = 1.96 - 8.48d\\\\8.48d = 1.96\\\\d = \frac{1.96}{8.48} \\\\d = 0.23\ m[/tex]
Thus, the distance traveled by the box before coming to rest is 0.23 m.
"Your question is not complete, it seems to be missing the following information";
A box is sliding up an incline that makes an angle of 15 degrees with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.40 m/s. How far does the box travel along the incline before coming to rest.
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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1i + b2j is added to vector ~A the resulting vector has no x component and a negative y component of 4.43 units. Find the value of
a. b1
b. b2
Answer:
a. 3.07 b. 1.26
Explanation:
Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j
Since A + B = -3.07i + 3.17j + b1i + b2j
= (-3.07 + b1)i + (3.17 + b2)j
So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j
Comparing components,
-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)
a. From (1), b1 = 3.07
b. From(2) b2 = 4.43 - 3.17 = 1.26
a traveling wave is described by equation y(x,t)=0.003(20x+200t) whereby y and x are measured in meters and t in second .what is the period of this wave?
Answer:
0.0314secs
Explanation:
The standard equation of a wave is expressed as;
y(x,t) = Asin(2πx/λ+2πft)
compare and contrast with the equation y(x,t)=0.003(20x+200t)
2πft = 200t
2πf = 200
f = 200/2π
f = 100/π
Since period T = 1/f
T = π/100
T = 3.14/100
T = 0.0314secs
hence the period of the wave is 0.0314secs
Please help!! There’s 10 points for it
Answer:
1200 J
Explanation:
option 1 should be the answer
1. Calculate the heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.750.000 joules of heat,
and its temperature changes from 32°C to 57°C.
Answer:
1.8 J/gºC
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 1500 g
Heat (Q) absorbed = 67500 J
Initial temperature (T₁) = 32 °C
Final temperature (T₂) = 57 °C
Specific heat capacity (C) =?
Next, we shall determine the change in temperature of the wood. This can be obtained as follow:
Initial temperature (T₁) = 32 °C
Final temperature (T₂) = 57 °C
Change in temperature (ΔT) =.?
ΔT = T₂ – T₁
ΔT = 57 – 32
ΔT = 25 °C
Finally, we shall determine the heat capacity of the wood. This can be obtained as follow:
Mass (M) = 1500 g
Heat (Q) absorbed = 67500 J
Change in temperature (ΔT) = 25 °C
Specific heat capacity (C) =?
Q = MCΔT
67500 = 1500 × C × 25
67500 = 37500 × C
Divide both side by 37500
C = 67500 / 37500
C = 1.8 J/gºC
Thus, the heat capacity of the wood is 1.8 J/gºC
whitch two options are forms of kinetic energy?
Answer:the witch has nothing to do with the problem
Explanation:
2. Denisse walks 8 km east in 2 hours.
• What is Denisse's average velocity?
Answer:
Probably 4
Explanation:
Help help help help please
Answer:
The sediment settled with the largest particles at the bottom and the smaller at the top.
I have 17 liters of air to a balloon the is 200 kelvin. If I take the balloon to a place where the temperature is 157 kelvin, what is the new volume of the balloon be
Answer:
The new volume of the ballon will be 13.345 L
Explanation:
Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure, by means of a constant of proportionality that is applied directly. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.
Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
It is possible to assume that you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
In this case:
V1= 17 LT1= 200 KV2= ?T2= 157 KReplacing:
[tex]\frac{17 L}{200 K} =\frac{V2}{157 K}[/tex]
Solving:
[tex]V2= 157 K*\frac{17 L}{200 K}[/tex]
V2= 13.345 L
The new volume of the ballon will be 13.345 L
16. Why does the number of carts matter when designing a roller coaster track? (Hint: PE = mass x gravity x height and KE = /2 mass x velocity ^2)
Choose 2 correct statements.
A. Adding carts increases the mass and decreases the total energy in the system.
B. Adding carts increases the mass and increases the total energy in the system.
C. Removing carts increases the mass and decreases the total energy in the system.
D. Removing carts decreases the mass and decreases the total energy in the system.
Answer:
Answer B. Adding carts increases the mass and increases the total energy in the system.
Explanation:
By adding carts, the mass of the system is larger, and therefore, both the potential energy and the kinetic energy of the system will increase, thus contributing to larger final velocities as the carts roll down the tracks.
The correct answer is therefore the one shown in answer B:
Adding carts increases the mass and increases the total energy in the system.
the bouncing back of sound when it hits ahard surface is called
Answer:
Reflection of sound
Explanation:
Sound waves bounce back from hard surface's.
A 15.0 kg bowling ball rolling at 3 m/s is stopped by a constant force of 11.2 newtons. Calculate the time the force must act on the bowling ball to stop it.
Answer:
4.01 seconds
Explanation:
Given that:
Mass of ball = 15kg
Initial velocity, u = 3m/s
Final velocity, v = 0
Force, F= 11.2 N
Change in velocity, dv = 3 - 0 = 3
Time force must act on the ball before stopping it:
Using the relation :
F = ma
a = (v - u) / t
Ft = m(v - u)
11.2 * t = 15 * 3
11.2t = 45
11.2t = 45
t = -+¯ 45 / 11.2
t = 4.01
t = 4 seconds.
A 1.40-kg ball bounces off a vertical wall. The ball approaches the wall at 9.70 m/s to the east and leaves the wall with the same speed. What is the change in momentum that the wall imparts to the ball
Answer:
The change in momentum is 0
Explanation:
Step one:
given data
mass of ball = 1.4kg
initial velocity of ball u = 9.7m/s
final velocity of ball v = 9.7m/s
Required
the change in momentum
Step two:
From the expression for momentum
P=mv
the change in momentum
Δp= mu-mv
Δp= 1.4*9.7-1.4*9.7
Δp= 13.58-13.58
Δp=0
There is no change in momentum
A container with a mass of 5 kg is lifted to a height of 8 m
It takes 392 joules of work to lift it.
It has 392 joules of gravitational potential energy up there.
Explain why a bolt of lightning is like the spark you might see when you touch a metal object and get a shock.
Answer:
Just like lightning, the spark you see is the discharge of static electricity that equalizes the charges. When you touch a metal object and get a shock, electrons are travelling in between you and the object to equalize the charges of the two objects. The light that is seen is the plasma created by electrons jumping between objects which heats the air surrounding them.
mike took less time to run the 100 meters than mitchell what does this tell us about his speed
An ice skater with rotational inertia I0 is spinning with angular speed omega subscript 0. She pulls her arms in, thereby increasing her angular speed to 4 omega subscript 0 . Her rotational inertia is then: a. 2 I0 b. I0 c. I0 /2 d. I0 /4 e. 4 I0
Answer:
d. I0 /4
Explanation:
Given that:
Initial angular velocity of the skater = [tex]\omega_o[/tex]
After pulling arms, angular velocity changes to = [tex]4 \omega _o[/tex]
Provided that no external torque is acting on skater;
Then:
[tex]\tau _{ext} = 0[/tex]
The torque [tex]\tau_{ext} = \dfrac{dL}{dt}[/tex]
where;
[tex]\tau _{ext} = 0[/tex]
It implies that:
[tex]\dfrac{dL}{dt} =0[/tex]
where;
L = constant
[tex]L_{initial} = L_{final}[/tex]
So;
[tex]I_i\omega_i = I_f \omega _f[/tex]
[tex]I_f =\dfrac{I_i\omega_i}{ \omega _f}[/tex]
[tex]I_f =\dfrac{I_o\omega_o}{ 4\omega _o}[/tex]
[tex]\mathbf{ I_f =\dfrac{I_o}{ 4}}[/tex]
Find the radioactivity of a 1 g sample of 226Ra given that [tex]t_{1/2}=1620[/tex] years and Avogadro's number = 6.023 × [tex]10^{23}[/tex].
Answer:
Explanation:
No of atoms of Ra in 1 g of sample = 6.023 x 10²³ / 226
N = 2.66 x 10²¹
disintegration constant λ = .693 / half life
half life = 1620 x 365 x 60 x 60 x 24 = 5.1 x 10¹⁰ s
disintegration constant λ = .693 / 5.1 x 10¹⁰
radioactivity dn / dt = λN
= (.693 / 5.1 x 10¹⁰ ) x 2.66 x 10²¹
= .3614 x 10¹¹ per sec
= 3.614 x 10¹⁰ / s
what does newton's first law describes
What type of mirror can produce both
converging and diverging rays?
Answer:
A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses.
Explanation:
hope this helps have a good night :)
Concave mirror can produce both converging and diverging rays.
What is concave mirror?A hollow spherical can be made into a mirror by cutting it into pieces, painting the exterior surface, and using the interior surface as the reflecting surface. Concave mirrors are this kind of mirror.
When object is placed inside focal length, convex mirror diverges the light beams and produces virtual image.
When object is placed outside focal length, convex mirror converges the light beams and produces real and diminish image.
Learn more about concave mirror here:
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Two small nonconducting spheres have a total charge of (a) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each
Answer:
q = 1 x 10⁻⁵ C = 10 μC
Explanation:
The repulsive force between the charges is given by Coulumb's Law:
[tex]F = \frac{kq_{1}q_{2}}{r^{2}}\\[/tex]
where,
F = Electrostatic Force = 12 N
k = Coulomb's Constat = 9 x 10⁹ Nm²/C²
r = distance between charges = 28 cm = 0.28 m
Since the values or charges are not given. We assume that both charges have same mahnitude. Therefore,
q₁ = q₂ = q = charge on each sphere = ?
Therefore,
[tex]12\ N = \frac{(9\ x\ 10^{9}\ Nm^{2}/C^{2})q^{2}}{(0.28\ m)^{2}} \\\\q^{2} = \frac{(12\ N)(0.28\ m)^{2}}{9\ x\ 10^{9}\ Nm^{2}/C^{2}}\\q = \sqrt{1\ x\ 10^{-10}\ C^{2}}\\[/tex]
q = 1 x 10⁻⁵ C = 10 μC
Calculate the wave length of the red light in air if the frequency of red light is 4.3×10^14hz. Given the velocity of light is 3×10^8m/s
Answer:
λ = 6.98 × 10^-7 m
Explanation:
Using the formula; λ = v/f
Where;
λ = wavelength (m)
v = speed of light (3 × 10^8m/s)
f = frequency (Hz)
Based on the provided information in this question, the frequency (f) of red light is 4.3×10^14hz while v = 3×10^8m/s.
λ = v/f
λ = 3×10^8 ÷ 4.3×10^14
λ = 0.698 × 10^(8-14)
λ = 0.698 × 10^-6
λ = 6.98 × 10^-7 m
3. Superman is flying on the sky, suddenly approaches an airplane. It is known that this airplane has 1,000,000 kg*m/s of momentum. Superman decides to push the airplane with a force of 1,200,000 N for .09 seconds. What is the impulse that superman gives to the plane? What is the new momentum of the airplane?
Answer:
Superman's delivered impulse : 108,000 kg m/s
New momentum of the airplane: 1,108,000 kg m/s
Explanation:
Recall that impulse can be estimated by multiplying the applied force times the duration of time the force was applied. Therefore, the impulse added by Superman was:
1,200,000 * 0.09 = 108,000 kg m/s
and then, the new momentum of the plane is the addition:
1000000 + 108000 = 1,108,000 kg m/s
The normal eye, myopic eye and old age
As we age, the lens of the eye hardens and accommodation no longer occurs: it is presbyopia.
Paradoxically, with age, a nearsighted person sees better than a "normal" person.
The aim of this exercise is to explain this paradox. The eye is modeled by a constant focal
length f′0 when the eye is presbyopic and by a screen (the retina) at the distance d from the
lens (d = 15 mm).
1. A normal, presbyopic eye sees an object at infinity in focus while a myopic eye sees an
object at a distance Dm from the eye (Dm = 15 cm).
What relation do we have in the two cases between f′0 and d?
2. A presbyopic person reads a newspaper placed 25 cm from his/her eyes. The radius r0 of
the pupil of the eye is 1 mm. Calculate the diameter of the spot on the retina image from a
log point, for a myopic eye and a normal eye. Conclude
Answer:
1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) the two diameters have the same order of magnitude and are very close to each other
Explanation:
You have some problems in the writing of your exercise, we will try to answer.
1) The equation to be used in geometric optics is the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where p and q are the distance to the object and the image, respectively, f is the focal length
* For the normal eye and with presbyopia
the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)
[tex]\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}[/tex]
f'₀ = 1.5 cm
this is the focal length for this type of eye
* Eye with myopia
the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm
1 / f = 1/15 + 1 / 1.5
1 / f = 0.733
f = 1.36 cm
this is the focal length for the myopic eye.
In general, the two focal lengths are related
f’₀ / f = 1.5 / 1.36
f’₀ / f = 1.10
The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit
a sin θ= m λ
the first zero occurs for m = 1, as the angles are very small
tan θ = y / f = sin θ / cos θ
for some very small the cosine is 1
sin θ = y / f
where f is the distance of the lens (eye)
y / f = lam / a
in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor
y / f = 1.22 λ / D
y = 1.22 λ f / D
where D is the diameter of the eye
D = 2R₀
D = 2 0.1
D = 0.2 cm
the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation
* normal eye
the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation
\frac{1}{f} = \frac{1}{p} + \frac{1}{q}
sustitute
[tex]\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}[/tex]
[tex]\frac{1}{f}[/tex]= 0.7066
f = 1.415 cm
therefore the diffraction is
y = 1.22 550 10⁻⁹ 1.415 / 0.2
y = 4.75 10⁻⁶ m
this is the radius, the diffraction diameter is
d = 2y
d_normal = 9.49 10⁻⁶ m
* myopic eye
In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm
[tex]\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}[/tex]
[tex]\frac{1}{f}[/tex]= 0.733
f = 1.36 cm
diffraction is
y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2
y = 4.56 10-6 m
the diffraction diameter is
d_myope = 2y
d_myope = 9.16 10-6 m
[tex]\frac{d_{normal}}{d_{myope}}[/tex] = 9.49 /9.16
\frac{d_{normal}}{d_{myope}} = 1.04
we can see that the two diameters have the same order of magnitude and are very close to each other
