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2 years ago

Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.

Physics

1 answer:

ivann1987 [24]2 years ago

3 0

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

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first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

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2 years ago

A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an

masya89 [10]

Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say $\omega$

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

$v=r\cdot \omega$

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

$a_c=\omega ^2\times r$

7 0

3 years ago

What is the magnitude of the force that lifts a 3-kilogram object straight upwards?

frez [133]

Any force of 29.4 Newtons or greater can do it.

4 0

2 years ago

Read 2 more answers

The spans of time that make up the Earth's geologic timescale are generally defined by the different types of

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<span>division of Earth's history into time units based largely on the types of life-forms that lived only during certain periods.</span>

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3 years ago

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0

3 years ago