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2 years ago

How is the kinetic energy of the particles that make up a substance different

1 answer:

6 0

All particles have energy, and the energy varies depending on the temperature the sample of matter is in, which determines if the substance is a solid, liquid, or gas. Solid particles have the least amount of energy, and gas particles have the greatest amount of energy

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A 26-kg sled is on a snow-covered slope. The coeﬃcients of friction between the sled’s runners and the snow are µs = 0.096 and µ

sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction $\mu _s=0.096$

coefficient of kinetic friction $\mu _k=0.072$

In order to move sled from rest we need to provide a force greater than static friction which is given by

$f_s=\mu mg=0.096\times 26\times 9.8=24.46 N$

After Moving Sled kinetic friction comes in to play which is less than static friction

$f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N$

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0

2 years ago

Who actually asked Abraham to sacrifice his son?

pochemuha

I think god did ??? I searched it up okay

7 0

2 years ago

Read 2 more answers

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

2 years ago

PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w

d1i1m1o1n [39]

Answer:

$F = 0.3\ Hz$

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

$F = \frac{1}{T}$

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

$F = \frac{1}{3s}$

$F = 0.333\ Hz$

$F = 0.3\ Hz$ --- Approximated

7 0

2 years ago

Read 2 more answers

A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio

Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

v² = u² + 2 a h

0 = 64² - 2 x 32 x h

64 h = 64²

h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0

2 years ago