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Nostrana [21]
3 years ago
15

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

op-up such that after hitting the bat, the baseball is moving at 53.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.20 ms. What is the average vector force the ball exerts on the bat during their interaction?
Physics
1 answer:
Luda [366]3 years ago
6 0

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

F=ma

F=\dfrac{mv}{t}

F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}

F₁ = 2900 N...........(1)

F=ma

F=\dfrac{mv}{t}

F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

F=F_1+F_2

F=(2900i+(-3493.18)\ N

F=(2900i-3493.18j)\ N

Hence, this is the required solution.

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Answer: A,B, and E

Explanation: Just checked I got them right:)

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3 years ago
Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d
denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

Time Period = 3 div * 5msec /div\\Time Perod = 15 msec

Since,

Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz

8 0
3 years ago
A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0
3 years ago
A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio
rjkz [21]

Hi there!

We can use Newton's Second Law:

\Sigma F = ma

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0
2 years ago
Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
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