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Delvig [45]
3 years ago
9

Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the

following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
Physics
1 answer:
damaskus [11]3 years ago
4 0

Explanation:

Acceleration. Angular acceleration: Is the rate of change of the angular velocity of a body with respect to time.

Force. Torque: Is also called rotational force, since an applied torque will change the rotational motion of a body.

Mass. Moment of inertia: It is the resistance that opposes a body to rotates.

Work. Work: In a rotational motion, the work is done by the torque.

Translational kinetic energy. Rotational kinetic energy: is the kinetic energy due to the rotational motion of a body.

Linear momentum. Angular momentum: Represents the quantity of rotational motion of a body.

Impulse. Angular impulse: Is the change in angular momentum of a body.

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Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
Tems11 [23]

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

F_{f} = -\mu N

Since we're talking about the same car, the value of N will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

7 0
3 years ago
Read 2 more answers
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

6 0
3 years ago
Two soccer players try to kick one soccer ball with the same force at the same time in opposite directions. Describe the force a
slamgirl [31]

Answer:

It will do nothing. The forces are balanced and the ball's motion will not change

7 0
3 years ago
What is the difference between compound, element and mixture?
defon
Compound; consists of atoms of two or more different elements bound together,can be broken down into a simpler type of matter (elements) by chemical means (but not by physical means) has properties that are different from its component elements, and always contains the same ratio of its component atoms.Mixtures; Note that a mixture:consists of two or more different elements and/or compounds physically intermingled, can be separated into its components by physical means, and often retains many of the properties of its components.
4 0
3 years ago
what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific
weqwewe [10]

Answer:

115.2^{\circ}C

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by \Delta T, according to the equation

Q=mC_s \Delta T

where C_s is the specific heat capacity of the substance.

In this problem, we have:

Q=250 \cdot 4.184 =1046 J is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

C_s = 0.175 J/gC is the specific heat capacity of gold

Solving for \Delta T, we find the change in temperature

\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}

And since the final temperature was

T_f = 175^{\circ}

The initial temperature was

T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C

3 0
3 years ago
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