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3 years ago

Describe the acceleration of your bicycle as you ride it from your home to the store

1 answer:

8 0

While you're going to the store, your acceleration changes. Some times it increases your overall speed sometimes it reduces it. Constant acceleration does not occur because it would mean that you would constantly accelerate and eventually go past the store. Even reduction of speed is a type of acceleration in physics. When you reach it, we can then calculate how much your velocity was on average and analyze how changing acceleration would've affected it.

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2 years ago

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2 years ago

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A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m

Feliz [49]

Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$ . The kinetic energies of father and son are $K_f\ and\ K_s$ . The mass of father and son are $m_f\ and\ m_s$

(a) According to given conditions, $K_f=\dfrac{1}{3}K_s$

And $m_s=\dfrac{1}{4}m_f$

Kinetic energy of father is given by :

$K_f=\dfrac{1}{2}m_fv_f^2$ .............(1)

Kinetic energy of son is given by :

$K_s=\dfrac{1}{2}m_sv_s^2$ ...........(2)

From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$ ..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

$v_s^2=4(v_f+1.5)^2$

Using equation (3) in above equation, we get :

$v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s$

(b) Put the value of $v_f$ in equation (3) as :

$v_s=7.09\ m/s$

Hence, this is the required solution.

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3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

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3 years ago

Evaluate (x +y)0 for x= -3 and y=5. 0 1 2 01

frosja888 [35]

Answer:2.01201

Explanation:

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3 years ago