All categories

2 years ago

If a ball has a velocity of 18 m/s, how far does it travel in 15 seconds

Physics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

The answer is

Explanation:

To find the distance when given the velocity and time we use the formula

<h3>distance = velocity × time</h3>

From the question

velocity of the ball = 18 m/s

time = 15 s

So the distance is

distance = 18 × 15

We have the final answer as

Hope this helps you

You might be interested in

If the density of an object is 5.2 g/cm3, and volume is 3.7 cm3, what is its mass

netineya [11]

Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3

6 0

3 years ago

Read 2 more answers

An object with a 25 µC charge is 0.54 m away from a second charged object. They experience a force of 3250 N. What is the charge

nekit [7.7K]

25 uc charge is 0.54 m away

6 0

2 years ago

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

2 years ago

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$