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3 years ago

What is the speed of a 1200 kg car moving with a kinetic energy of 15 kJ?

Physics

1 answer:

Pepsi [2]3 years ago

7 0

Answer:5m/s

Explanation:

Kinetic energy=ke=15kj=15x1000j=15000j

Mass=1200kg

Speed=√(2ke/m)

Speed=√(2x15000/1200)

Speed=√(30000/1200)

Speed=√(25)

Speed=5m/s

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an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni

stiv31 [10]

The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N

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2 years ago

Science Mixtures Question( giving brainly, thanks, and rank of 5 stars.)

vazorg [7]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq

wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / $\sqrt{(R)^{2} + (Xl -Xc)^{2} }$

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

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3 years ago

a cannonball is fired with a speed of 76 m/s from the top of a cliff. It strikes the plane below with a speed of 89 m/s. if we n

RSB [31]

Answer:

we use the formula,

$v {}^{2} = u {}^{2} + 2gh$

$89 {}^{2} = 76 {}^{2} + 2(10)h$

$h = (89 {}^{2} - 76 {}^{2} ) \div 20$

h= 107 m

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3 years ago

Except when necessary for takeoff and landing, what is the minimum safe altitude required for a pilot to operate an aircraft ove

MAVERICK [17]

Except when necessary for takeoff and landing, <span>the minimum safe altitude required for a pilot to operate an aircraft over other than congested area is an altitude of 1000 ft above the highest obstacle within a 2000 ft horizontal radius of the aircraft.

It is also good to know that apart from taking off and landing, the aircraft must not operate at a distance less than 500 ft from any person, vessel, structure or vehicle.</span>

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2 years ago