Ask question

Ask question

All categories

2 years ago

8

A biker accelerates from 22m/s to 62m/s. His acceleration was 10m/s2. How long did it take to speed up? (select the correct equa

tion)

1 answer:

xenn [34]2 years ago

5 0

Answer:

Option 4

Explanation:

You might be interested in

_______is a speed in a certain direction.

Law Incorporation [45]

D. velocity
Velocity depends on speed and direction

6 0

2 years ago

Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving

Papessa [141]

Answer:

The final velocity of the object after 2 seconds is 30 m/s

Explanation:

Given;

constant downward acceleration, a = 10 m/s²

initial velocity of the object falling down, v = 10 m/s

time of fall, t = 2 s

The final velocity of the object is given by;

v = u + at

where;

v is the final velocity

v = 10 + (10)(2)

v = 10 + 20

v = 30 m/s

Therefore, the final velocity of the object after 2 seconds is 30 m/s

3 0

2 years ago

Which configuration would produce an electric current? A) Rotate a coil of copper wire between two magnets. B) Connect a wire be

FrozenT [24]

the answer is (A) the movement of the magnet relative to the coil

4 0

3 years ago

Read 2 more answers

How strongly the planet you're on pulls on you is your

ladessa [460]

Is the answer you are looking for Gravity? Gravity is what pulls us down to earth.

4 0

2 years ago

Read 2 more answers

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago