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GalinKa [24]
3 years ago
9

At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc

ed staff member knows that the deceleration of a car when skidding is -40.52 m/s2. What can the staff member estimate for the original speed of the race car if it came to a stop during the skid?
Physics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

27.1 m/s

Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

Using third equation of motion,

V^2 = U^2 + 2aS

Since the car is decelerating, the final velocity V = 0

Substitute all the parameter into the equation above,

0 = U^2 - 2 * 40.52 * 9.06

U^2 = 734.22

U = \sqrt{734.22}

U = 27.096

U = 27.1 m/s  approximately

Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

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<span>B. It stays the same</span>
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What would be the answer for this and how?
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Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

k = \frac{F}{\Delta x}\\

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

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k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

7 0
2 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
3 years ago
A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he
Oksanka [162]

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

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7 0
3 years ago
he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1
meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

5 0
3 years ago
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