Which of the following is a result of sea floor spreading?

A body travel according to the law x(t)=2t+3t3 calculate the accelaration of body at t=2s

faust18 [17]

Answer:

28m/s

Explanation:

Given law,

⇒ x(t) = 2t + 3t³

Where, t = 2s

Then,

⇒ x(2) = 2(2) + 3(2)³

⇒ 4 + 3(8)

⇒ 4 + 24

⇒ 28 m/s

Acceleration is 28m/s

4 0

3 years ago

• List two conditions needed for transfer of heat by conduction

Murljashka [212]

Answer:

Difference in temperature

Contact between the bodies

7 0

3 years ago

Read 2 more answers

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

2 years ago

A toy car sits motionless at the top of a ramp. What will happen to the car if a balanced force acts upon it?

dexar [7]

Answer:

kinetic energy because when the car moves again it will be faster then it was before

5 0

2 years ago

5. You are skateboarding on top of a small hill and take a break. You have 100 J of potential energy. If

Sholpan [36]

Given the potential energy of 100 J, the height of the hill is 0.128 m (12.8 cm)

Explanation:

The gravitational potential energy of an object is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

In this problem, we have

U = 100 J is the potential energy of the person

$g=9.8 m/s^2$ is the acceleration of gravity

m = 80 kg is the mass of the person

Solving for h, we find at what height the person is:

$h=\frac{U}{mg}=\frac{100}{(80)(9.8)}=0.128 m = 12.8 cm$

Learn more about gravitational potential energy here:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0

2 years ago