All categories

3 years ago

What is the kinetic energy of a 55-kilogram skier traveling at 9.0 meters per second?

1 answer:

8 0

<span>The formula to get the Kinetic Energy is KE=1/2mv^2 where m=mass in kilogram and v=velocity of the object. From you given, we substitute it to the variables. This is the solution: KE=1/2mv^2 KE=1/2(55kg)(9.0m/s)^2 KE=1/2(4455) KE=2227.5J or it can be written in scientific notation as 2.2275x10^3J or closest to your choices is 2.2x10^3J.</span>

You might be interested in

A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0\,\text m4.0m4, point, 0, start text, m, end

kirill [66]

Answer:

0.903 seconds

Explanation:

To find how many seconds the acorn fall, we can use the formula for distance travelled with constant acceleration:

D = Vo*t + a*t^2/2,

where D is the distance travelled, Vo is the inicial speed, t is the time and a is the acceleration.

In our problem:

Vo = 0,

a = g = 9.81 m/s2,

D = 4 meters.

So, we can solve the equation to find the time:

4 = 0*t +9.81*t^2/2

4.905*t^2 = 4

t^2 = 4/4.905 = 0.8155

t = 0.903 seconds

8 0

3 years ago

What is a creative name of a roller coaster about the solar system

GenaCL600 [577]

Solar roller is a cool name

7 0

3 years ago

during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled

wlad13 [49]

45mph is the answer if you do the math right

6 0

3 years ago

2 Physic Questions For 20 Points ✨

Anna [14]

Jupiter Cannot Become A Star.
Jupiter Is The Fastest Spinning Planet In The Solar System.
The Clouds On Jupiter Are Only 50 km Thick.

8 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

3 years ago