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3 years ago

A person hits a tennis ball with a mass of 0.058 kg against a wall.

Physics

1 answer:

Alla [95]3 years ago

6 0

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

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At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the oppo

Viefleur [7K]

Answer:

The average acceleration during the 6.0 s interval was -27 m/s².

Explanation:

Hi there!

The average acceleration is defined as the change in velocity over time:

a = Δv/t

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity

t = elapsed time

The change in velocity will be:

Δv = final velocity - initial velocity

Δv = -74 m/s - 87 m/s = -161 m/s

(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)

Then the average acceleration will be:

a = Δv/t

a = -161 m/s / 6.0 s

a = -27 m/s²

The average acceleration during the 6.0 s interval was -27 m/s².

8 0

3 years ago

A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running west with a speed of 3.6 m/s. Assum

MariettaO [177]

Answer:

The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West

Explanation:

given information:

mass of fullback, $m_{x}$ = 92 kg

speed of full back, $v_{x}$ = 5.8 to south

mass of lineman, $m_{y}$ =110 kg

speed of lineman, $v_{y}$ = 3.6

according to conservation energy,

assume that the collision is perfectly inelastic, thus

initial momentum = final momentum

$P_{ix}$ = $P_{x}$ '

m₁v₁ = (m₁+m₂) $v_{x}$ '

$v_{x}$ ' = m₁v₁/(m₁+m₂)

= (92) (5.8)/(92+110)

= 2.64 m/s

$P_{iy}$ = $P_{y}$ '

m₂v₂ = (m₁+m₂) $v_{y}$ '

$v_{y}$ ' = m₁v₁/(m₁+m₂)

= (110) (3.6)/(92+110)

= 1.96 m/s

thus,

$v$ ' = √ $v_{x}$ '²+ $v_{y}$ '²

= 3.3 m/s

then, the direction of the two players is

θ = 90 - tan⁻¹( $v_{y}$ '/ $v_{x}$ ')

= 90 - tan⁻¹(1.96/2.64)

= 53.4° South of West

7 0

3 years ago

Which is not standing in the way of astronomers getting a good view of distant stars? A. some stars are too far away for our tel

Naya [18.7K]

Answer: D. There is a lot of light pollution on earth

Explanation: The light pollution on Earth has nothing to do with the stars in the sky

3 0

3 years ago

If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?

nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

$h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}$

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.

4 0

2 years ago

Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:

Deffense [45]

Answer:

a) $T=0.01s$

b) $T=0.001s$

c) $T=0.00001s$

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

$T=\frac{1}{f}$

Therefore

For

$T=100 Hz$

$T=\frac{1}{100}$

$T=0.01s$

For

$F=1kHz$

$T=\frac{1}{1000}$

$T=0.001s$

For

$F=100kHz$

$T=\frac{1}{100*100}$

$T=0.00001s$

6 0

3 years ago