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Alchen [17]
3 years ago
11

The molecules of a substance become more closely packed and move more quickly.

Physics
1 answer:
andrey2020 [161]3 years ago
4 0
When the substance are moved close together and they move more quickly they get compressed.
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Coming to the bottom of a mountain, a skier moving with speed v collides with a barrier and is brought to a stop in an amount of
IRISSAK [1]

Answer:

Explanation:

Impulse of a force is measured by force x time or F X t

Impulse also equals change in momentum or

F x t = m v₂ - m v₁

The given case is as follows

in the first case

F x t = mv - o = mv

F = mv / t

in the second case

F₁ x 4 t = mv

F₁ = 1/4 x mv /t

F₁ = F / 4

option a) is correct .

iii )

In the last case

F₂ X t = m v/2 -0

F₂ = 1/2 x mv / t

= 1/2 x F

F₂ = F/2

Option e ) is correct.

8 0
2 years ago
A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i
anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

  • It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

       I = \frac{1}{2}*m*r^{2}  = \frac{1}{2}*0.400 kg*(0.0865m)^{2}  = 1.5e-3 kg*m2

3 0
3 years ago
The smallest living part that makes up all living things is a
Arturiano [62]

Answer:

cells

Explanation:

they are the smallest living organisms

5 0
3 years ago
My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
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