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WITCHER [35]
3 years ago
11

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

nd. The radius of the hammer's path is 2.0 meters.
Physics
1 answer:
Semenov [28]3 years ago
6 0

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

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The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 k
nekit [7.7K]

Answer:

d=79.9m

Explanation:

From the question we are told that:

coefficient of static friction \mu=0.38

Velocity v=87.9=>24.41667m/s

Generally the equation for Conservation of energy is mathematically given by

 \mu*mgd = 0.5 m v^2

 d=\frac{0.5*24.42^2}{0.38*9.8}

 d=79.9m

3 0
3 years ago
You are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceilin
Ivan

Answer: The elevator must be accelerating.

Explanation:

As the tension force is opposing to the the force of gravity on the load which is hung vertically, and the tension force can adopt any value in order to comply with Newton's 2nd law, if the tension force is less than the force due to gravity, this means that all system is not in equilibrium, so it must be accelerating.

If we assume that the downward is the positive direction, we can write:

mg - T = ma

If T = 0.9 mg, ⇒ mg (1-0.9) =0.1 mg = m a ⇒a = 0.1 g , in downward direction.

5 0
3 years ago
Why are quartz useful
Sladkaya [172]
The ability to convert voltage to and from mechanical stress.
8 0
3 years ago
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
3 years ago
QUESTION 3 ( MARKS]
horrorfan [7]

Answer:

392 N

Explanation:

Draw a free body diagram of the rod.  There are four forces acting on the rod:

At the wall, you have horizontal and vertical reaction forces, Rx and Ry.

At the other end of the rod (point X), you have the weight of the sign pointing down, mg.

Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.

Sum of the moments at the wall:

∑τ = Iα

(T sin θ) L − (mg) L = 0

T sin θ − mg = 0

T = mg / sin θ

Given m = 20 kg and θ = 30.0°:

T = (20 kg) (9.8 m/s²) / (sin 30.0°)

T = 392 N

7 0
4 years ago
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