Ask question

Ask question

All categories

3 years ago

11

A tennis ball with a speed of 11.3 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the oppo

site direction with a speed of 7.2207 m/s. if the ball is in contact with the wall for 0.00803 s, what is the average acceleration of the ball while it is in contact with the wall? take "toward the wall" to be the positive direction. answer in units of m/s 2 .

1 answer:

Arlecino [84]3 years ago

4 0

To calculate the average acceleration of the ball we use the formula,

$a= \frac{v_{f} -v_{i}}{t}$

Here, $v_{f}$ is the final velocity of the ball and $v_{i}$ is the initial velocity of the ball t is the time in contact with the wall.

Given $v_{i} = 11.3 m/s$ towards the wall and $v_{f} = 7.2207 m/s$

away from the wall and $t=0.00803 s$ .

Substituting these values in above formula , we get

$a=\frac{(-7.2207 m/s)-11.3 m/s}{0.00803 s}$

Here $v_{f}$ is negative because ball is moving away from the wall.

$a= - 2306 .4 \ m/s^{2}$

Therefore, average acceleration of the ball is $- 2306 .4 \ m/s^{2}$ (away from the wall).

You might be interested in

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

2 years ago

How does air pressure change with altitude?

charle [14.2K]

Air pressure changes with altitude because of issues related to gravity. Molecules have more weight the closer they are to the Earth and more of them move to lower elevations as a result; this causes increased pressure because there are more molecules in number and proximity. Conversely, air at higher elevations has less weight, but also forces pressure on those layers below it, resulting in the molecules closer to the Earth supporting more weight, increasing the pressure

3 0

3 years ago

If it requires 7.0 j of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re

SVEN [57.7K]

4.6 j more. To get this take 7 and multiply it by 3.5 to get 24.5 take the x which is what you’re looking for and multiply it by the 2.1 to get 2.1x. Take 24.5 and divide it by 2.1 x and get 11.6. Subtract 11.6 by 7 and get 4.6

8 0

3 years ago

A copper wire has a diameter of 2.097 mm. what magnitude current flows when the drift velocity is 1.54 mm/s? take the density of

frozen [14]

By using drift velocity of the electron, the current flow is 7.20 ampere.

We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as

v = I / (n . A . q)

where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.

From the question above, we know that

d = 2.097 mm

r = (0.002097 / 2) m

v = 1.54 mm/s = 0.00154 m/s

ρ = 8.92 x 10³ kg/m³

q = e = 1.6 x 10¯¹⁹C

Find the atom density

n = Na x ρ / Mr

where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).

n = 6.022 x 10²³ x 8.92 x 10³ / 0.635

n = 8.46 x 10²⁷ /m³

Find the current flows

v = I / (n . A . q)

0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)

0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)

I = 7.20 ampere

For more on drift velocity at: brainly.com/question/25700682

#SPJ4

6 0

2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago