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algol13
3 years ago
8

Where are some of Earth's youngest rocks found?

Physics
1 answer:
Over [174]3 years ago
4 0
No the answer is C at the seafloor spreading centers
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What type of acceleration does an object moving in a circular path with constant speed experience?
Marysya12 [62]

An object moving in a circular path has centripetal acceleration. <em>(A)</em>

7 0
3 years ago
Read 2 more answers
A person, with a mass of 50.0 kg, stands on a weighing scale in a lift which is moving
svet-max [94.6K]

Answer:

D

Explanation:

Because I just had that answer

6 0
3 years ago
A 1.2 nf parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nf when the gap is filled
Damm [24]

Answer:

2.5

Explanation:

The capacitance of a parallel-plate capacitor filled with dielectric is given by

C=kC_0

where

k is the dielectric constant

C_0 is the capacitance of the capacitor without dielectric

In this problem,

C_0=1.2 nF is the capacitance of the capacitor in air

C=3.0 nF is the capacitance with the dielectric inserted

Solving the equation for k, we find

k=\frac{C}{C_0}=\frac{3.0 nF}{1.2 nF}=2.5

3 0
3 years ago
Read 2 more answers
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats
Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

7 0
3 years ago
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