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2 years ago

Ice floats on water.

Physics

1 answer:

OleMash [197]2 years ago

5 0

Answer:

Yeah ice floats on water.

Observation

Example in those areas were ice is found like Antarctica ice is found on top of water.

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What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7

vitfil [10]

The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.

<h3>What kind of gravitational pull does the moon have on the planet?</h3>

On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.

<h3>What does the Earth's center's gravitational pull feel like?</h3>

Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).

<h3>Where is the Earth's and the moon's gravitational centre?</h3>

It is around 1700 kilometres below Earth's surface.

To know more about gravitational force visit:-

brainly.com/question/12528243

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6 0

1 year ago

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

Anon25 [30]

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>

This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

F is the applied force
v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

7 0

2 years ago

It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)

Dividing by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 = (6/5)*x² - 2*x - 24/5

Using quadratic formula

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}$

$x = \frac{2 \pm 5.2}{2.4}$

$x_1 = \frac{2 + 5.2}{2.4}$

$x_1 = 3$

$x_2 = \frac{2 - 5.2}{2.4}$

$x_2 = -1\; 1/3$

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

Please help ASAP please

Alex_Xolod [135]

Answer:

Explanation:

6 0

2 years ago

Ice floats on water.​

Ice floats on water.