Ask question

Ask question

All categories

3 years ago

13

10 POINTS AND BRAINLIEST!!!!

2 answers:

Maurinko [17]3 years ago

7 0

The answer is earth.

Darya [45]3 years ago

7 0

In the Earth it will be returns to energy to the atmosphere and water vapor.

Answer will be bolded.

You might be interested in

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

$F_{branch} + F_{gravity}=F_{centripetal}$

force of gravity = - mg

$F_{branch}=F_{centripetal}-F_{gravity}$

= $\dfrac{mv^2}{r} + mg$

= $m(\dfrac{3.4^2}{0.6} +9.8)$

=9 x 29.067

= 261.6 N

upward force acting = 261.6 N

7 0

3 years ago

When light passes from water into air, the light bends away from the normal. Question 4 options: True False

kykrilka [37]

Answer:

True

Explanation:

8 0

3 years ago

A muon is traveling at 0.988

AlexFokin [52]

As per Einstein's relation of relativity

$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

here we know that

$m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg$

$m_0 = 1.88 \times 10^{-28} kg$

now here we know that

$v = 0.988 c$

now from above equation mass of the muon is given as

$m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}$

$m = 1.22 \times 10^{-27} kg$

now for the momentum of muon we can use

$P = mv$

$P = 1.22 \times 10^{-27} \times 0.988(3 \times 10^8)$

$P = 3.62 \times 10^{-19} kg m/s$

so above is the momentum of muon

6 0

3 years ago

Which of the Moon’s positions will cause the highest or spring tides?

Bess [88]

The correct answer is A and C

During a Spring Tide, it will be horizontal, and a Neap tide is vertical

Brainliest plz!

8 0

3 years ago

Read 2 more answers

A 0.75-kg ball falls vertically downward from a height of 55.0 m and rebounds upward. if the ball reaches a height of 30.0 m and

Nikitich [7]

Impuls I is given by:
$I = \delta mv = |F|t$
where $\delta mv$ is the change in momentum, $|F|$ is the average force and t is the time.

Solve the equation for the force F:
$|F| = \frac{\delta mv}{t} = \frac{m(v_f - v_i)}{t}$

Energy should be conserved, so the velocities will be:
$\frac{1}{2}mv^2 = mgh \\ v = \sqrt{2gh}$

Combining both equations:
$|F| = \frac{m( \sqrt{2g(h_f + h_i)} )}{t}$
where $h_f = 30m, h_i = 55m, m = 0.75kg, t = 0.0025s$

4 0

3 years ago