All categories

3 years ago

Another athlete using a different spring exerts an average force of 400 N to enable her

Physics

1 answer:

babymother [125]3 years ago

6 0

Answer:84Nm

Explanation:

force=400N

Distance=0.210m

Workdone=force x distance

Workdone=400 x 0.210

Workdone=84Nm

You might be interested in

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

What is the most important source of water vapor in the atmosphere?

guapka [62]

The answer is Oceans. It is the most important source of water vapor in the atmosphere. <span>Given the huge amount of water they have and their huge surface areas, naturally the bulk of water which evaporates and enters the atmosphere is from oceans.</span>

8 0

3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

3 years ago

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e

Natali5045456 [20]

Answer:

a) $F = 660.576\,N$ , b) $a_{c} = 6.953\,\frac{m}{s^{2}}$ , c) $v \approx 7255.423\,\frac{m}{s}$ , $\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$ , d) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

$F = G\cdot \frac{m\cdot M}{r^{2}}$

$F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}$

$F = 660.576\,N$

b) The centripetal acceleration of the satellite is:

$a_{c} = \frac{660.576\,N}{95\,kg}$

$a_{c} = 6.953\,\frac{m}{s^{2}}$

c) The speed of the satellite is:

$v = \sqrt{a_{c}\cdot R}$

$v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}$

$v \approx 7255.423\,\frac{m}{s}$

Likewise, the angular speed is:

$\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}$

$\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$

d) The period of the satellite's rotation around the Earth is:

$T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)$

$T \approx 1.821\,h$

6 0

3 years ago

Since nuclear fusion in the sun creates energy from matter, why doesn't it violate the law of conservation of energy?

alexandr402 [8]

Since nuclear fusion in the sun creates energy from matter, Einstein's formula E=mc² states that matter and energy are equal.

To find the answer, we have to know more about the nuclear fusion.

<h3>What is nuclear fusion?</h3>

We are aware that the sun can achieve nuclear fusion by the fusion of hydrogen atoms.
These atoms need to get closer to one another in order to fuse.
Since both protons inside each nucleus are positively charged, they attempt to repel one another as they get closer to one another.
If this issue cannot be solved, nuclear fusion in the sun cannot occur.

Thus, we can conclude that, since nuclear fusion in the sun creates energy from matter, Einstein's formula E=mc² states that matter and energy are equal.

Learn more about nuclear fusion here:

brainly.com/question/25663405

#SPJ4

3 0

2 years ago