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3 years ago

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

nd the 80.00 m radius circular track. What is the frequency of the race car's circular motion?

1 answer:

koban [17]3 years ago

3 0

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

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The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall

ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is also acting downward as it is creating friction between ball and air molecules, so creating hindrance in upward motion

External force to throw ball upward

So

Net Force = Upward force - Air friction - Weight

Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

So

Net Force = Weight - Air friction

Since ball is going downward, so weight is greater than net force which is in this case is air friction which is pulling ball upward.

4 0

4 years ago

A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc

Digiron [165]

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

$p_1 = p_2$

$m_1u_1+m_2u_2 = m_1v_1+ m_2v_2$

Here,

$m_{1,2} =$ Mass of each object

$u_{1,2} =$ Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

$m_1u_1 = (m_1+m_2)v_f$

Rearranging to find the velocity,

$v_f = \frac{m_1u_1}{ (m_1+m_2)}$

Replacing we have that,

$v_f = \frac{(332)(2.3)}{ (332+19.5)}$

$v_f = 2.17 m/s$

Therefore the velocity of the shark immediately after it swallows the tank is $2.17m/s$

4 0

4 years ago

As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.

scZoUnD [109]

Answer:22.36 m/s

Explanation:

Given

the diameter of loop d=50 m

the radius of loop r=25 m

At the top position, we can write,

weight and Normal reaction combination will provide the centripetal force i.e.

$R+W=\frac{mv^2}{r}$

$R=W\quad \quad [\text{apparent weight =Actual weight}]$

$2W=2mg=\frac{mv^2}{r}$

$v=\sqrt{2gr}$

$v=\sqrt{2\times 10\times 25}$

$v=22.36\ m/s$

3 0

3 years ago

A force of 6 N acts on a 33 kg object for 9 seconds. What is the objects change in velocity?

GuDViN [60]

Acceleration = change in velocity/time
By F = ma,
6 = 33 x change in velocity / 9
change in velocity = +1.636 m/s

4 0

3 years ago

A golden retriever is sitting in a park when it sees a squirrel. The dog starts running, exerting a constant horizontal force of

oksano4ka [1.4K]

Answer:

284.8 kgm/s

Explanation:

Impulse: This can be defined as the product of force and time of a body. The S.I unit of impulse is N.s mathematically.

Impulse = Force × time

Change in momentum: This is the product of the mass of a body and its change in velocity. The unit of change in momentum is kgm/s.

Mathematically,

momentum = mass×change in velocity

Deduction from newton's second law of motion,

Impulse = change in momentum

Therefore,

Change in Momentum = Force×time

ΔM = F×t................. Equation 1

Where F = force = 89 N, t =time = 3.2 s.

Substitute into equation 1

ΔM = 89×3.2

ΔM = 284.8 kgm/s

Thus the change in momentum = 284.8 kgm/s

8 0

3 years ago