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mars1129 [50]
2 years ago
15

Find the force of gravity between a cubic meter of water (1000kg) and the Sun. The Sun's mass is 1.99 x10^30 kg and is 1.50 x10^

11 m away.
A) 17.0 N
B) 5.90 N
C) 2.0 N
D) 47.20 N
Physics
1 answer:
alisha [4.7K]2 years ago
5 0

Answer:

F = 5.9 N

Explanation:

Given that,

Mass of water, m₁ = 1000 kg

Mass of the Sun, m_2=1.99\times 10^{30}\ kg

Distance between two bodies, r=1.5\times 10^{11}\ m

We need to find the force of gravity between water and the Sun. The formula for the gravitational force between two bodies is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1000\times 1.99\times 10^{30}}{(1.5\times 10^{11})^2}\\\\F=5.89\ N

or

F = 5.9 N

Hence, the correct option is (b).

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You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine
DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

F_x = F cos \theta

with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

F_f = \mu mg

where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N

8 0
3 years ago
It would be really helpful if u help me solving this question. PLEASE!!!
sweet [91]

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice 

water

heatproof mat 

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

7 0
3 years ago
If you exert a force of 100.0 N to lift a box a distance of 0.5 m, how much work do you do? 200 J 400 J 50 J 26 J
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What phenomena provides <br> evidence of electric fields in the atmosphere
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Which prediction of weather is the most accurate for the next two days if there is an occluded front over the area?
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The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.

During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.

The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.

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