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liberstina [14]
4 years ago
6

A car is traveling on a level road at a constant rate of speed. What additional force is necessary to bring the car into equilib

rium? zero. greater than the normal force times the coefficient of static friction. the normal force times the coefficient of kinetic friction. equal to the normal force times the coefficient of static friction.​
Physics
1 answer:
iragen [17]4 years ago
5 0

Answer: Zero.

Explanation:

By the first Newton's law, we know that:

every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

Now, we know that the car is moving with constant speed, then there is no net force acting on the car, which means that the car is already in equilibrium.

Then if we add one force to the situation, the car will not be anymore in equilibrium.

The correct option is zero.

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The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro
ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

4 0
3 years ago
A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0
4 years ago
How do i solve this?
kenny6666 [7]

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

7 0
3 years ago
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
4 years ago
To view an enlarged upright image of an object through a simple magnifier, where must the object be located?.
Andre45 [30]

Answer:within the focal length of the lens, provided the focal length is shorter than the near point distance.

Explanation:Hope it helps

6 0
3 years ago
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