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Ksivusya [100]
4 years ago
10

A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In t

erms of only the givens vo and h, derive expressions for the final speed of the ball and the elapsed time of flight.

Physics
1 answer:
yuradex [85]4 years ago
4 0

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

v^{2}=u^{2}+2gh

And the time is

v = u + gt

t=\frac{v-u}{g}

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S the work done on Amanda's car while speeding up (i) greater than, (ii) less than, or (iii) the same as the work done on Bertha
ankoles [38]

Answer:

Explanation:

Mass of amanda and bertha car = Ma = Mb

Initial velocity of amanda, ua = 10 m/s

Final velocity of amanda, va = 20 m/s

Initial velocity of bertha, ub = 20 m/s

Final velocity of bertha, vb = 30 m/s

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Ea = Ma × 1/2 × (20^2 - 10^2)

= Ma × 150

= 150 Ma J

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Eb = Mb × 1/2 × (30^2 - 20^2)

= Mb × 250

= 250 Mb J

Option ii. Less than. Thus this is because the workdone by amanda car (150Ma) is less than the workdone in berthas car (250Mb).

B.

Impulse, p = force × time

p = Mass × change in velocity

pa = Ma × va - Ma × ua

= Ma × (20 - 10)

= 10 × Ma

= 10 Ma

p = Mass × change in velocity

pa = Mb × vb - Mb × ub

= Mb × (30 - 20)

= 10 × Mb

= 10 Mb

Option iii. The same as.

The impulse as seen above is the same in amanda car (10Ma) as the same with bertha (10Mb)

7 0
4 years ago
HELP ASAP!!! Some bands hate playing in school gyms because sound waves easily reflect off the walls and floor. What could you d
Virty [35]

Answer:

I would recommend placing something on the ground or on the walls of your gym to damped the echos.

Explanation:

6 0
4 years ago
What is the equivalent resistance of the circuit?
podryga [215]

Answer:

Option D. 5.45 Ω

Explanation:

From the question given above, the following data were obtained:

Resistance 1 (R₁) = 10 Ω

Resistance 2 (R₂) = 20 Ω

Resistance 3 (R₃) = 30 Ω

Voltage (V) = 120 V

Equivalent resistance (R) =?

The equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/10 + 1/20 + 1/30

Find the least common multiple (lcm) of 10, 20 and 30. The result is 60

Divide 60 by each of the denominators and multiply by their numerators respectively. We have:

1/R = (6 + 3 + 2)/60

1/R = 11/60

Invert

R = 60/11

R = 5.45 Ω

Thus, the equivalent resistance in the circuit is 5.45 Ω

8 0
3 years ago
At a processing facility outside of Detroit, a 7.0 kg box of goat cheese, initially at rest, is given a push up a smooth, inclin
Firlakuza [10]

Answer:

d = 5.9 m

Explanation:

During the initial push, there are two forces acting on the box along  the ramp: the component of gravity force along the ramp (directed to the bottom of the ramp), and the pushing force (up the ramp).

As we have as a given, the time during which the pushing force is applied, we can use Newton's 2nd Law, expressed in its original form, as follows:

Fnet *Δt = Δp = m* (vf-v₀) (1)

Fnet, in this case, is the difference between Fapp, and the projection of Fg along the ramp, which is equal to Fg times the sinus of  the angle of the ramp with respect to the horizontal.

We choose as positive, the direction up the ramp, so we can write the follwing equation:

⇒ Fnet = Fapp - Fg*sin θ = 140 N - (7kg*9.8m/s²*sin 30º) = 140 N - 34.3 N

⇒ Fnet = 105.7 N

Replacing in (1):

105.7 N * 0.5 s = 7 kg* vf (v₀=0, as the box starts from rest)

Solving for vf:

vf = 105.7 N* 0.5 s / 7 kg = 7.6 m/s

When the push ends, the only force remaining along  the ramp, is the component of Fg that we have already obtained, that will cause the box to have a deceleration, which we can find out aplying Newton's 2nd Law, as follows:

m*g*sin 30º = m*a

As we have defined as positive direction the one up the ramp, a will be negative (as it is slowing down the box) , and can be calculated as follows:

a = -34.3 N / 7 kg = -4.9 m/s²

As this value is constant, we can use any kinematic equation in order to get  the distance traveled, farther the point where it disappeared the influence of the pushing force:

vf² - v₀² = 2*a*d

As we know that finally the box will come momentarily at rest (before falling under the influence of  gravity) , we have vf =0:

⇒ -v₀² = 2*a*d

For this part, v₀, is just the value for vf, that we got above:

v₀= 7.6 m/s

⇒ -(7.6)² =2*(-4.9 m/s²)*d

Solving finally for d (the answer we are looking for):

d = (7.6)² (m/s)² / 2*4.9 m/s² = 5.9 m

8 0
3 years ago
Which of the following is the best example of kinetic energy being transformed into potential energy
poizon [28]

A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).

3 0
4 years ago
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