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1 year ago

To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration

2 answers:

7 0

In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.

Jlenok [28]1 year ago

4 0

According to Newton's first law, the law of inertia, an object's velocity will not change unless it is acted upon by an outside force. This means that objects at rest tend to stay at rest, and objects in motion tend to stay in motion.

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A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t

baherus [9]

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return , it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

6 0

2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago

A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

Lisa [10]

Answer:

Vertical distance= 3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time= 0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0

2 years ago

At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci

kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²