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Sergeeva-Olga [200]
4 years ago
14

When the temperature of an object increases, what happens to the kinetic energy of the particles in the object?

Physics
1 answer:
d1i1m1o1n [39]4 years ago
3 0
When the average kinetic energy of the particles in the object increases, we sense and measure that change as an increase in the object's temperature.
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A circketer lowers his hands <br>hile catching a ball . why?​
zzz [600]
To absorb the force of the ball
3 0
3 years ago
A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
Georgia [21]

Answer:

Wp = 1,272 J

Wf = -940.8 J

Wn = 0

Wg = 0

Explanation:

  • Applying the definition of work, as the product of the component of the  force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
  • As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
  • We can get the work done by the force applied P, simply as follows:

        W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)

  • Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
  • The frictional force , can be written as follows:

        F_{fr} = \mu k * F_{n} (2)

        where μk = coefficient of kinetic friction = 0.24

         Fn = Normal Force

  • In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
  • Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
  • Replacing these values in (2), we get:

       F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N

  • Applying the definition of work, we can get the work done by the frictional force, as follows:

        W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ =  W_{ffr} = -940.8 N

4 0
3 years ago
A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
postnew [5]

Answer:

20 m/s^2

Explanation:

Use the formula Vf=Vi+at

Vf=100

Vi=0

a=?

t=5

Then solve for (a)

7 0
3 years ago
A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
3 years ago
22. A race car accelerates from 0.0 m/s to 5 m/s with a displacement of 2.5 m. What is the
yanalaym [24]

Answer:

Option A

5 m/s^{2}

Explanation:

From fundamental equation of motion

v^{2}=u^{2}+2as where v is the final velocity, u is the initial velocity, a is the acceleration of the body and s is the displacement.

Since the initial velocity is zero

v^{2}=2as

Making acceleration, a the subject of the formula then

a=\frac {v^{2}}{2s}

Substituting 5 m/s for v and 2.5 m for s then

a=\frac {5^{2}}{2\times 2.5}=5 m/s^{2}

6 0
3 years ago
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