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Pie
3 years ago
6

Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

mph in three seconds, looking nonchalant as he passes the officer. how far does he travel while braking?
Physics
1 answer:
barxatty [35]3 years ago
5 0
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
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A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t
baherus [9]

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return ,  it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

6 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surfa
Pavel [41]

Answer:

C. Up, equal to the can's weight

Explanation:

You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surface of a lake. What is the direction and amount of force the water exerts on it?

A. Zero

B. Down, equal to the can's weight

C. Up, equal to the can's weight

D. Not enough information is given

from the principle of flotation which states that a

When a body displaces a weight of water equal to its own weight, it floats. : A floating object displaces a weight of fluid equal to its own weight. ... Archimedes' principle equates the buoyant force to the weight of the fluid displaced.

the upthrust (this is the upward vertical force exerted on an object in fluid)in the water equals the weight of the body in water it floats.

7 0
3 years ago
a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi
e-lub [12.9K]

Answer:

en español por

favor para entender un poco más

4 0
3 years ago
A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round
Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s

v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s

Now we find the centripetal acceleration which is given by

a_c=\frac{v^2}{r}

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

a_t = \frac{v-u}{t}

where

t = 17.0 s

Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

6 0
3 years ago
Read 2 more answers
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