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Nataly [62]
3 years ago
12

A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?

Physics
1 answer:
pochemuha3 years ago
7 0
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
The energy that generates wind comes from what source?
Rudiy27

Answer:

we can say that wind energy is due to

D) Severe thunderstorms

Explanation:

As we know that wind energy is converted into kinetic energy of wind mills

This kinetic energy of wind mill is then converted into electrical energy using turbine

now we can consider here  energy conservation theory that energy is only converted from one form to other form

it neither be destroyed nor be created but it can transfer from one form to other form

So here we can say that wind energy is due to

D) Severe thunderstorms

3 0
3 years ago
A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of
garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0
3 years ago
What two factors determine the amount of friction there will be between objects?
barxatty [35]

Answer:

the two factors are the mass of the objects and the coefficient of friction between them

Explanation:

internet :)

6 0
3 years ago
A 235.0 g metal block absorbs 2.44 × 103 J of heat to raise its temperature by 35 K. What is the specific heat of the metal? Sho
andrew-mc [135]
When an object absorbs an amount of energy equal to Q, its temperature raises by \Delta T following the formula
Q=m C_s \Delta T
where m is the mass of the object and C_s is the specific heat capacity of the material.

In our problem, we have Q=2.44 \cdot 10^3 J, m=235.0 g and \Delta T=35 K, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
C_s =  \frac{Q}{m \Delta T}= \frac{2.44 \cdot 10^3 J}{(235.0 g)(35 K)}=0.297 J g^{-1} K^{-1}
3 0
3 years ago
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