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dsp73
3 years ago
15

What is the minimum speed needed to fire a champagne cork a distance of 11m?

Physics
1 answer:
Crank3 years ago
8 0

To start with solving this problem, let us assume a launch angle of 45 degrees since that gives out the maximum range for given initial speed. Also assuming that it was launched at ground level since no initial height was given. Using g = 9.8 m/s^2, the initial velocity is calculated using the formula:

(v sinθ)^2 = (v0 sinθ)^2 – 2 g d

where v is final velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m

Rearranging to find for v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>

<span>v0 = 10.383 m/s</span>

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
PLEASE HELP COME UP WITH 5 MULTIPLE CHOICE QUESTIONS FROM HEALTH RELATED FITNESS AND SKILL RELATED QUESTIONS
Helga [31]

Answer:

1. What is not considered a cardiovascular workout?

A. Jogging

B. Sit ups

C. Jump Ropes

D. Burpees

Explanation:

I am sorry, I am in middle school and I only have one question.

5 0
2 years ago
A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we
Alex_Xolod [135]

Answer:

 w = 2w₀     the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

       I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

         L₀ = L

       I₀ w₀ = I w

       I₀ w₀ = I₀ / 2 w

       w = 2w₀

therefore the angular velocity of man doubles

7 0
3 years ago
One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e
frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

E= \Delta mc^{2}

where \Delta m = change in mass

c = speed of light = 3 \times 10 ^{8}m/s

Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0
3 years ago
A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely  falling object.

3 0
2 years ago
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