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nordsb [41]
3 years ago
10

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates at 1.5m/s2? A. 51.5 N B. 4

8.5 N C. 75 N D. 33.3 N
Physics
1 answer:
const2013 [10]3 years ago
3 0
Use the Second Law of Newton, which states that Net Force = mass times acceleration

F = m * a

F = 50 kg * 1.5 m/s^2 = 75 N

Answer: option C
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3 years ago
An object has a coefficient of static friction of 0.3 and a normal force of 30 N. Find the force of static friction.
gladu [14]

Answer:

9N

Explanation:

static friction=normal force x coefficient of static friction

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A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Pachacha [2.7K]

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

6 0
3 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
3 years ago
What is the main function of chloroplasts?
dangina [55]

Answer:

C

To convert sunlight into usable sugars

(

Explanation:

6 0
3 years ago
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