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nika2105 [10]
3 years ago
9

A toboggan approaches a snowy hill moving at 12.4 m/s. The coefficients of static and kinetic friction between the snow and the

toboggan are 0.400 and 0.290, respectively, and the hill slopes upward at 44.0 degrees above the horizontal. What is the acceleration of the toboggan going up the hill? and the acceleration after it has reached its highest point and it sliding downhill?
Physics
1 answer:
Harrizon [31]3 years ago
3 0

Answer:

a) Acceleration while the toboggan is going uphill = 8.85 m/s²

b) Acceleration of the toboggan is going downhill = 4.76 m/s²

Explanation:

Resolving the normal reaction into vertical and horizontal component

Nₓ = mg sin θ

Nᵧ = mg cos θ

The frictional force on the toboggan = Fr = μmg cos θ (note that μ = μ(k) = the coefficient of kinetic friction)

Doing a Force balance on the x component taken in the axis parallel to the inclined plane.

The net force that has to accelerate the toboggan has to match the frictional force acting downwards of the plane (In the opposite direction to motion) and x-component of the Normal reaction

ma = Nₓ + Fr

ma = (mg sin θ) + (μmg cos θ)

a = g[(sin 44°) + (0.29 cos 44°)

a = 9.8(0.6947 + 0.2086) = 8.85 m/s²

b) While coming downhill,

The frictional force is acting uphill now (In the direction opposite to motion),

The force balance is

ma = (mg sin θ) - (μmg cos θ)

a = g[(sin 44°) - (0.29 cos 44°)]

a = 9.8(0.6947 - 0.2086)

a = 4.76 m/s²

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3 years ago
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A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli
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Answer:

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = v_{f}=?

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}

Substituting the values we get

2\times 14 + 5\times 0 =(2+5)\times v_{f}

v_{f}=\dfrac{28}{7}=4\ m/s

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

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amid [387]
Washington DC and new Mexico
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Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a
MariettaO [177]

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

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3 years ago
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