Ask question

Ask question

All categories

4 years ago

7

Two people on skateboards, Brian and Amy, push off from each other. Brian has more mass than Amy. Who experiences a larger force

?

1 answer:

amm18124 years ago

4 0

Amy because she is lighter, so the more force will be put on her

You might be interested in

A baseball with a mass of 0.15 kg is moving at a speed of 40 m/s. What is

Afina-wow [57]

Answer:

120 J

Explanation:

KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J

6 0

3 years ago

Read 2 more answers

A cube with sides of length 2cm has a mass of 7.36g . calculate the density of the cube

kondor19780726 [428]

Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3

7 0

3 years ago

What is the period if the block’s mass is doubled? note that you do not know the value of either m or k , so do not assume any p

Oksanka [162]

The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.

The time period T of the block with mass M attached to a spring of spring constant K is given by,

T = 2π(√M/K).

Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,

T' = 2π(√2M/K)

Putting T = 2π(√M/K) above,

T' =√2T

So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.

To know more about time period of mass, visit,

brainly.com/question/20629494

#SPJ4

5 0

1 year ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

or

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

3 years ago

How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2

elena-s [515]

The heat transferred by the steam to the skin is given by
$Q=m L_v$
where
m is the mass of the steam
$L_v$ is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
$m=25.0 g=0.025 kg$
while the latent heat of vaporization of the steam is
$L_v = 2.256 \cdot 10^6 J/kg$
Substituting into the previous formula, we find the heat transferred to the skin:
$Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J$

8 0

4 years ago