WILL GIVE BRAINLIEST!!!

Water near the poles would most likely be stored as

solniwko [45]

Hey there!

Answer: Glaciers

Water near the poles would most likely be stored as glaciers. Glaciers are slow moving rivers that are a buildup of ice and snow.

Thank you!

5 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

A 2000-kg railway freight car coasts at 4.4 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

ruslelena [56]

Answer:

933. 3kg

Explanation:

We are given that

Mass,m=2000 kg

Initial speed,u=4.4 m/s

We have to find the maximum mass of grain if the speed of loaded freight car must not go below 3.0 m/s.

Final speed,v=3 m/s

By conservation of momentum

Initial momentum=Final momentum

$mu=m'v$

Substitute the values

$2000\times 4.4=m'(3)$

$m'=\frac{2000\times 4.4}{3}$

$m'=2.93\times 10^3 kg$

Mass of freight loaded car=2933.3 kg

Mass of grains=2933.3-2000=933.3 kg

Hence, the maximum mass of grains that it can accept=933.3 kg

7 0

3 years ago

A long straight wire carries a 40.0 A current in the +x direction. At a particular instant, an electron moving at 1.0  107 m/s

ser-zykov [4K]

Answer:

$F=-1.28\times 10^{-16}\ N$

Explanation:

Given that,

Current flowing in the wire, I = 40 A (+x direction)

Speed of the electron, $v=10^7\ m/s$ (+y direction)

Distance from the wire, r = 0.1 m

Let F is the electric force on the electron. It is given by :

$F=qvB\ sin\theta$

Here, $\theta=90$

$F=qvB$

Here, $B=\dfrac{\mu_oI}{2\pi r}$

$F=qv\dfrac{\mu_oI}{2\pi r}$

$F=-1.6\times 10^{-19}\times 10^7\times \dfrac{4\pi\times 10^{-7}\times 40}{2\pi \times 0.1}$

$F=-1.28\times 10^{-16}\ N$

So, the force on the electron at this instant is $-1.28\times 10^{-16}\ N$ . Hence, this is the required solution.

5 0

3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50 kg mass and 0.090 m radius. The bearings of the pulle

madreJ [45]

Answer:

1.8 m/s²

36 N

34.8 N

Explanation:

For the monkey :

m₁ = mass of monkey = 4.50 kg

T₁ = Tension force in the rope on monkey's side

a = acceleration

From the force diagram, force equation for the motion of monkey is given as

m₁ g - T₁ = m₁ a

(4.50 x 9.8) - T₁ = 4.5 a

T₁ = 44.1 - 4.5 a eq-1

For the bunch of bananas :

m₂ = mass of bunch of bananas = 3 kg

T₂ = tension force in the rope on the side of banana

From the force diagram, force equation for the motion of bananas is given as

T₂ - m₂ g = m₂ a

T₂ - (3 x 9.8) = 3 a

T₂ = 29.4 + 3 a eq-2

m = mass of the pulley = 1.50 kg

r = radius of the pulley = 0.090 m

α = angular acceleration of pulley = a/r

Torque equation for the pulley is given as

(T₁ - T₂ )r = I α

(T₁ - T₂ )r = I (a/r)

T₁ - T₂ = (0.5 m r²) (a/r²)

T₁ - T₂ = (0.5) ma

using eq-1 and eq-2

44.1 - 4.5 a - (29.4 + 3 a) = (0.5) ma

44.1 - 4.5 a - (29.4 + 3 a) = (0.5) (1.50) a

a = 1.8 m/s²

Using eq-1

T₁ = 44.1 - 4.5 a

T₁ = 44.1 - 4.5 (1.8)

T₁ = 36 N

using eq-2

T₂ = 29.4 + 3 a

T₂ = 29.4 + 3 (1.8)

T₂ = 34.8 N

7 0

3 years ago