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2 years ago

1. Consider the position vs time graph at right.

Physics

1 answer:

Lemur [1.5K]2 years ago

5 0

Answer:

Labor Day Weekend--No School

9/1/2021

Hello Students!

There will be no School on Friday Sept 3 and Monday Sept 6

All centers will be closed on Friday 9/3 and Monday 9/6

School will resume on TUESDAY, Sept 7

Enjoy your long weekend!

Explanation:

Labor Day Weekend--No School

9/1/2021

Hello Students!

There will be no School on Friday Sept 3 and Monday Sept 6

All centers will be closed on Friday 9/3 and Monday 9/6

School will resume on TUESDAY, Sept 7

Enjoy your long weekend!

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An unwary football player collides with a padded goalpost while running at a velocity of 9.50 m/s and comes to a full stop after

Amanda [17]

Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

8 0

2 years ago

A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made

frozen [14]

At time $t$ seconds, the mass has angular speed

$\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

and hence linear speed

$v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

After 8 s, its linear speed is

$v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}$

and has centripetal acceleration with magnitude

$a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}$

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

$F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}$

5 0

1 year ago

Energy can come from electricity, but it can also come from water false true

Hitman42 [59]

Answer:

the answer is truth

Explanation:

6 0

2 years ago

Read 2 more answers

Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

2 years ago

how would the velocity of the book change if the applied force were equal to the sliding friction force

Inessa [10]

Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is 9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton

3 0

3 years ago