Ask question

Ask question

All categories

3 years ago

6

A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The

bat acts on the ball for 0.019 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N Â· s.

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

$\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s$

So the change in momentum of the ball would be the product of its velocity change and its mass

$\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s$

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

You might be interested in

A bicycle going 10 m/s on a flat road demonstrates __________ energy.

vampirchik [111]

Then it demonstrate KINETIC ENERGY.

5 0

3 years ago

PROJECTILE MOTION FOR TWO Rocks-VELOCITY-TIME GRAPHS

erastova [34]

Answer:

nob

nobody knows i tried it doesnt work like that ask for help

Explanation:

5 0

3 years ago

Youssef and Sophia are driving in two different cars and not paying attention to their surroundings. Youssef approaches an inter

Reptile [31]

Answer:

I think E--

Explanation:

3 0

2 years ago

6. A rifle is aimed horizontally at shoulder height (1.5 meters above the ground) at a target bullseye 700 meters away. The bull

devlian [24]

a) The bullet will not reach the target

The motion of the bullet follows a parabolic path. First, we have to determine the time it takes for the bullet to reach the ground. We can do it by considering the vertical motion only, which is a free-fall motion, so we can use the equation

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

where, taking downward as positive direction,

s = 1.5 m is the vertical displacement of the bullet

u = 0 is its initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.5)}{9.8}}=0.553 s$

So, the bullet lands 0.553 s after being shot.

The bullet is fired horizontally at a speed of

$v_x = 1000 m/s$

So, the horizontal distance covered during this time is

$d=v_x t = (1000)(0.553)=553 m$

And since the target is 700 m away, the bullet will not reach it.

b) 553 m

As we stated in the previous part, the bullet takes

t = 0.553 s

To land to the ground.

Also, it travels at

$v_x = 553 m/s$

Therefore, it lands on the ground at a distance of

$d=v_x t=(1000)(0.553)=553 m$

4 0

3 years ago

A 3-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath with k = 0.13 W/m·K, and the wire/sheath interfac

Phoenix [80]

Answer:

$Q_{max} = 6,3928 W/m$

Explanation:

The thermal resistances in this case act as series resistances, so to get the total resistance you need to add al resitances:

$R_{tot} = R_{contact} + R_{conductive} + R_{convective}$

To get the contact resistence per unit length we have to multiply the contact resistance by the perimeter of the contact area:

$P_{wire-sheath} = \pi * 0,003m = 9,4248 * 10^{-3} m$

$R_{contact} = \frac{3*10^{-4} m2 K/W}{9,4248 * 10^{-3} m} = 0,03183 m K/W$

For the conductive resistance on the sheath:

$R_{conductive} = \frac{thickness}{k*P_{wire-sheath}} = \frac{ 0,002m}{9,4248 * 10^{-3} m * 0,13 W/mK} = 1,63 mK/W$

Convective resistance:

The perimeter of the auter surfice is:

$P_{out-sheath} = \pi * 0,007m = 2,199* 10^{-2} m$

$R_{convective} = \frac{1}{Convection coeficient * P_{out-sheath} } = \frac{ 1}{15 W/m^{2}K * 2,199* 10^{-2} m} = 3,031 mK/W$

Total Resistance:

$R_{tot} = 4,6928 mK/W$

Heat transfer:

$Q = dT/R_{tot}$

$dT = T_{wire} - T_{amb} = 323 K - 293 K = 30 K$

$Q_{max} = \frac{30K}{4,6928 mK/W} = 6,3928 W/m$

4 0

3 years ago