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Veseljchak [2.6K]
2 years ago
5

If the Earth’s mass decreased, how would the gravity between the Sun and Earth change?

Physics
2 answers:
irina [24]2 years ago
8 0

2- the gravity would decrease

andriy [413]2 years ago
8 0
(2)- decrease in gravity
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Illustrate a body of mass 5.0kg pulled by horizontal force F . if the body accelerates 2.0ms² and experience a frictional force
Natasha2012 [34]

Explanation:

a. Net force is mass times acceleration (Newton's second law).

∑F = ma

∑F = (5.0 kg) (2.0 m/s²)

∑F = 10 N

b. The net force is the sum of the individual forces.

10 N = F − 5 N

F = 15 N

c. Friction force here is mgμ.

mgμ = 5 N

(5.0 kg) (10 m/s) μ = 5 N

μ = 0.1

3 0
3 years ago
The sound wave passes from the sea-water into the air. State what happens to; the speed of the waves
dusya [7]
Answer. Explanation: Frequency of the sound decreases and the speed of sound becomes 346m/s from near about 1500 m/s.
6 0
3 years ago
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A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,
taurus [48]

Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

8 0
3 years ago
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A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict
Elan Coil [88]

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

8 0
3 years ago
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A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an
allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

K_i + U_i = K_f + U_f

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×v_f² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×v_f²

v_f² = 77.93/(7/10 × 4.75)

v_f ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, v_f ≈ 4.84 m/s.

3 0
2 years ago
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