All categories

3 years ago

What is the displacement of a person that ran 15 minutes if he or she runs at an average velocity of 16 km/hr west

Physics

1 answer:

a_sh-v [17]3 years ago

4 0

displacement = velocity × time

= 16 × 15

= 240

You might be interested in

If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?

balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

$U = K$

$\frac{kQe}{R} = \frac{1}{2}mv^2$

$\frac{2kQe}{mR} = v^2$

now take square root of both sides

$v =\sqrt{\frac{2kQe}{mR}}$

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

$k = 9 * 10^9$

5 0

3 years ago

Read 2 more answers

the engine of a car of amass of 2000 kg produced a force of 15000N find the acceleration of the car

weqwewe [10]

I believe the answer would be 7.5 m/s^2

3 0

3 years ago

How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)

maria [59]

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

$C=4186 J/g^{\circ}C$ (specific heat capacity of water)

$\Delta T=10.0^{\circ}C$

Therefore, the amount of energy needed is

$Q=(0.050)(4186)(10)=2093 J$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

4 0

3 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

3 years ago

What is the effect on the force of gravity between two objects if the mass of one object doubles?

Leni [432]

Then the force will also be doubled

6 0

3 years ago