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Blababa [14]
3 years ago
9

A 513 g ball strikes a wall at 14.7 m/s and rebounds at 11.3 m/s. The ball is in contact with the wall for 0.038 s. What is the

magnitude of the average force acting on the ball during the collision? Answer in units of N.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

consider the velocity of the ball towards the wall as negative and away from the wall as positive.

m = mass of the ball = 513 g = 0.513 kg

v₀ =  initial velocity of the ball towards the wall before collision = - 14.7 m/s

v = final velocity of the ball away from the wall after collision = 11.3 m/s

t = time of contact with the wall = 0.038 sec

F = average force acting on the ball

using impulse-change in momentum equation , average force is given as

F = m (v - v₀)/t

inserting the values

F = (0.513) (11.3 - (- 14.7))/0.038

F = 351 N


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At a country music festival, a band is playing at the end of a crowded
Anna11 [10]

Answer:

t=4.86s

Explanation:

To find the wavelength you use the following formula:

v=\lambda f

v: speed of sound = 343m/s

f: frequency = 400Hz

λ: wavelength of the sound

By doing λ the subject of the formula and replacing the values of f and v you obtain:

\lambda=\frac{v}{f}=\frac{343m/s}{400Hz}=0.85m

Now, to calculate the time that sound takes to reach the last row you use:

t=\frac{d}{v}

t: time

d: distance to the last row = 1947m

t=\frac{1947m}{400Hz}=4.86s

hence, the time is 4.86s

4 0
4 years ago
The space shuttle releases a satellite into a circular orbit 630 km above the Earth.
solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

  • \displaystyle g=\frac{GM}{r^2}

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

  • \displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}

Remove units to make the equation easier to read.

  • \displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}

Multiply the numerator out.

  • \displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}

Add the terms in the denominator.

  • \displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}

Simplify this equation.

  • \displaystyle g=8.11045189 \ \frac{m}{s^2}

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

  • \displaystyle a = \frac{v^2}{r}

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

  • \displaystyle 8.11045189 \ \frac{m}{s^2}  = \frac{v^2}{7008100 \ m}  

Remove units to make the equation easier to read.

  • \displaystyle 8.11045189   = \frac{v^2}{7008100}

Multiply both sides by 7,008,100.

  • 56838857.89=v^2

Take the square root of both sides.

  • v=7539.154985

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0
3 years ago
Cindy runs 500 meters every morning. It takes her 2 hours to complet
marta [7]

Answer:

250 / 2.5*10^2

Explanation:

7 0
2 years ago
A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th
erica [24]
QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
6 0
3 years ago
Using this information...
Pepsi [2]

19.2\:\text{m/s}

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)

\:\:\:\:\:= 19.2\:\text{m/s}

6 0
3 years ago
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