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2 years ago

11 An unstretched spring is 12,0 cm long. A load of 5.0N stretches it to 15.0cm. How long will it be under a load

Physics

1 answer:

Rom4ik [11]2 years ago

4 0

Answer: Approximately 22 cm

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Explanation:

The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.

Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).

3.0 cm = (3.0)/100 = 0.03 m

Apply Hooke's Law to find the spring constant k

F = -kx

5.0 = -k*(0.03)

k = -(5.0)/(0.03)

k = -166.667 approximately

Now we must find the displacement x when F = 15 newtons

F = -kx

-kx = F

x = F/(-k)

x = -F/k

x = -15/(-166.667)

x = 0.089 approximately

x = 0.1

The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm

So the spring will be stretched to about 12cm+10cm = 22 cm

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2 years ago

If the force on the tympanic membrane (eardrum) increases by about 1.4 N above the force from the atmospheric pressure, the memb

Sveta_85 [38]

Answer:

10.58 ft

Explanation:

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Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2

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Pressure at depth, P = h x d x g

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2 years ago

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene

Nataliya [291]

Question in proper order

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic energy term is called the kinetic energy of the center of mass.

You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell.

It is important to understand the applicability of the formula

$Ktot=Kr+Kt$

Which of the following conditions are necessary for the formula to be valid?

a. The velocity vector $v$ must be perpendicular to the axis of rotation

b.The velocity vector $v$ must be perpendicular or parallel to the axis of rotation

c. The moment of inertial must be taken about an axis through the center of mass

Answer:

Option c

Explanation:

$K_{total} = K_{rotational}+K_{translational}$

The first two conditions are untrue, this is because, you can have rotation in any direction and translation in any direction of any collection of masses. Rotational and translational velocities of masses do not depend on each other

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3 years ago

Read 2 more answers

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NemiM [27]

Usain bolt is 43km/h
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5 0

2 years ago

a cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. when a 10 kg cannonball is fired to the left at a s

marishachu [46]

Answer:

0.5 m/s

Explanation:

From Newton's Third law of motion,

Momentum of the cannon = momentum of the flatcar.

mv = MV........................ Equation 1

Where m = mass of the cannon, v = velocity of the cannon, M = mass of the flatcar, V = Velocity of the flat car.

make V the subject of the equation

V = mv/M................... Equation 2

Given: m = 10 kg, v = 50 m/s, M = 1000 kg

Substitute into equation 2

V = 10×50/1000

V = 500/1000

V = 0.5 m/s

Hence the speed of the flatcar = 0.5 m/s

6 0

3 years ago