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2 years ago

1.

Physics

1 answer:

drek231 [11]2 years ago

8 0

Answer:

2500m

Explanation:

given,

initial velocity(u) = 5m/s

final velocity (v) = 0 m/s ( as he/she stopped riding after 1000 secs)

time = 1000 secs

distance(s)=?

now,

s=( u+v ) ×t /2

=(5+0)×1000/2

=2500 m

hence, he /she travels 2500m far

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3. What do we call the ONLY part of the electromagnetic spectrum that we can

otez555 [7]

Answer:

Visible Light

wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns

1 A unit = 10^-10 m

1 mμ = 10^-9 m

6 0

2 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

6 0

3 years ago

Read 2 more answers

A trolley going down an inclined plane has an acceleration of 2cm/s^2 What will be its velocity

Snowcat [4.5K]

Answer:

$V_{f}$ = 6 cm/s

Explanation:

<u><em>Given:</em></u>

Acceleration = a = 2 cm/s²

Time = t = 3s

Initial Velocity = $V_{i}$ = 0 cm/s

<u><em>Required:</em></u>

Velocity = $V_{f}$ = ?

<u><em>Formula:</em></u>

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

<u><em>Solution:</em></u>

2 = $\frac{V_{f}-0}{3}$

=> $V_{f}$ = 2*3

=> $V_{f}$ = 6 cm/s

5 0

3 years ago

The distance between earth and mars is 51 G meters. What is the minimum RTT for a link between Earth and Mars if speed of light

beks73 [17]

Answer:

340 seconds = 5.667 minutes

Explanation:

As we know, S = v t or t = S / v (S = 51 x 10^9 m and v = 3 x 10^8 ms^-1)

So, t = 51 x 10^9 / 3 x 10^8 = 17 x 10^1 = 170 s

For a RTT estimation, the time span will be doubled of one way propagation for transmission and receive delay.

The over all round trip time will be = 170 x 2 = 340 seconds = 5.667 minutes

4 0

2 years ago

In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against

Viefleur [7K]

Answer:

24.07415 rpm

Explanation:

$\mu$ = Coefficient of friction = 0.63

v = Velocity

d = Diameter = 4.9 m

r = Radius = $\frac{d}{2}=\frac{4.9}{2}=2.45\ m$

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

Here the frictional force balances the rider's weight

$f=\mu F_n$

The centripetal force balances the weight of the person

$\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s$

Velocity is given by

$v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s$

Converting to rpm

$2.52104\times \frac{60}{2\pi}=24.07415\ rpm$

The minimum angular speed for which the ride is safe is 24.07415 rpm

4 0

2 years ago