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Thepotemich [5.8K]
3 years ago
11

What causes one plate to sink under another 1 Mass 2 Shape 3 Density

Physics
1 answer:
solong [7]3 years ago
3 0
It is either the mass or density, I believe it is 3 density tho but it could be mass good luck:)
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Suppose a particle is accelerated through space (no gravity) by a 10 N force. Suddenly the particle encounters a second force of
iren2701 [21]

please dont mind me just looking for points

3 0
3 years ago
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

c = Speed of light = m/s

Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
3 years ago
Which statements accurately describe mechanical waves? Check all that apply.
kotegsom [21]

Answer:

A

Explanation:

E. An ocean wave moving through water is an example of a mechanical wave

e.g sound waves wave on a rope or string

and Ans a is also correct

4 0
2 years ago
A bike travels at 7.5 m/s along a straight road, whereas a car travels at 10.0 m/s along the same road and in the same direction
Ilya [14]

Answer:

t = 8.3 s

Explanation:

As we know that

velocity of bike = 7.5 m/s

velocity of car is 10 m/s

deceleration of car is 0.75 m/s^2

part a)

velocity of bike with respect to car is given as

v_r = 7.5 - 10 = -2.5 m/s

acceleration of bike with respect to car is given as

a_r = 0 - (-0.75) = 0.75 m/s^2

now the distance of the bike with respect to car is given as

d = v_i t + \frac{1}{2}at^2

5 = (-2.5) t + \frac{1}{2}(0.75)t^2

t = 8.3 s

Part b)

3 0
3 years ago
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