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Yanka [14]
3 years ago
14

P.E QUESTION

Physics
1 answer:
vredina [299]3 years ago
4 0
The answer is 5 maybe
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A 38,500 kg sphere is located 2.55 m from a 15,400 kg sphere. What is the gravitational force, rounded to the nearest thousandth
kkurt [141]

The gravitational force between the two spheres is 0.006 N

Explanation:

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

For the two spheres in this problem, we have

m_1 = 38,500 kg

m_2 = 15,400 kg

r = 2.55 m

Substittuting into the equation, we find

F=(6.67\cdot 10^{-11})\frac{(38,500)(15,400)}{2.55^2}=0.006 N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

3 0
4 years ago
When air pressure decreases, the mercury in a barometer _____.
siniylev [52]

Answer:

The mercury in the barometer will go down as there is less air pressing down on the bulb of the barometer to push mercury up.

Explanation:

7 0
3 years ago
What is the magnetic force on a 200 cm length of (straight) wire carrying a current of 30 A in a region where a uniform magnetic
Bezzdna [24]

Answer:

F = 0.112 N

Explanation:

To find the magnitude of magnetic force on the wire, you use the following formula:

|\vec{F}|=|i\vec{L}\ X\ \vec{B}|=iLBsin\theta   (1)

L: length of the wire = 200cm = 0.2m

i: current in the wire = 30 A

B: magnitude of the magnetic field = 0.055 T

θ: angle between the directions of L and B = 20°

You replace the values of L, i, B and θ in the equation (1):

|\vec{F}|=(30A)(0.2m)(0.055T)sin(20\°)=0.112N

hence, the magnetic force on teh wire is 0.112N

3 0
3 years ago
What part of earth systems interact to form a storm like this hurricane near Florida
Akimi4 [234]
Hurricanes form from interactions between the atmosphere and the oceans. Hope it helps.
7 0
3 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
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