Identify which gas had the greatest volume change.

A metal sphere with a mass of 0.026 kg rolls along a friction-less surface at 3.7 m/s and strikes a stationary sphefe having a m

zubka84 [21]

The force applied to the second ball by the first ball is 6.734 × 10^-4 N.

<h3>What is impulse of force?</h3>

The impulse of force is defined as the sum of the average force and the duration it is applied.

If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.

Given that: mass of a metal sphere: m = 0.026 kg.

Initial speed of the sphere: u = 3.7 m/s.

When the sphere stops completely, its change in momentum = mu - 0

= 0.026×3.7 N-s.

= 0.0962 N-s.

As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied to the second ball =

change in momentum of 1st ball × time of contact

= 0.0962 × 0.007 N

= 0.0006734 N

= 6.734 × 10^-4 N.

Hence, the force applied to the second ball is 6.734 × 10^-4 N.

Learn more about impulse force here:

brainly.com/question/29787329

#SPJ1

7 0

1 year ago

Some trees have red leaves all year long. You know that plants with green leaves make their food for themselves. How do you thin

marshall27 [118]

The same way as the trees with green leaves. Through photosynthesis :)

6 0

3 years ago

Read 2 more answers

you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

Ivanshal [37]

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

$r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m$

So its area must be

$A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2$

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

$v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s$

7 0

3 years ago

What is weight if m is mass and g is gravitational force.

Nataly_w [17]

Weight is a force, and is defined as mass times acceleration. So (g is acceleration and m is mass):
$F_g=mg$

That's your answer.

7 0

3 years ago

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, the

ICE Princess25 [194]

Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by $\Delta L_{thick}$ and $\Delta L_{thin}$ , respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e $\frac{1}{2}$ option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

$E = \frac{\sigma}{\epsilon}$

Where $\sigma$ is the stress = $\frac{Force}{Area}$

$\epsilon$ is the strain = $\frac{\Delta L}{L}$

Making Strain the subject

$\epsilon = \frac{\sigma}{E}$

now in this question we are that the same tension was applied to both wires so

$\frac{\sigma}{E}$ would be constant

Hence

$\frac{\Delta L}{L} = constant$

for the two wire we have that

$\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}$

Looking at young modulus formula

$E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }$

$E * \frac{\Delta L }{L} = \frac{F}{A}$

$A * \frac{\Delta L}{L} = \frac{F}{E}$

Now we are told that a comprehensive force is applied to the wire so for this question

$\frac{F}{E}$ is constant

And given that the length are the same

so

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}$

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

Assuming that

$A_2 = 2 A_1$

So

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}$

$\frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}$

From the question the length are equal

$\Delta L_{thin} =2 \Delta L_{thick}$

So

$\frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}$

Hence the ratio is less than 1

6 0

3 years ago