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Kisachek [45]
3 years ago
7

Please help fast!!

Physics
1 answer:
daser333 [38]3 years ago
4 0

The answer is C..........      

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Name two objects that have a high density.
Sonbull [250]

Answer:

Iron and stone

Explanation:

4 0
3 years ago
A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and
MariettaO [177]

Answer:

f=140\ N

Explanation:

Given:

  • mass of the object on a horizontal surface, m=50\ kg
  • coefficient of static friction, \mu_s=0.3
  • coefficient of kinetic friction, \mu_k=0.2
  • horizontal force on the object, F=140\ N

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

F_s=\mu_s.N

where:

N= normal force of reaction acting on the body= weight of the body

F_s=0.3\times (50\times 9.8)

F_s=147\ N

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

f=140\ N

8 0
3 years ago
what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3
Oduvanchick [21]
V=m×d
m=v/d
m=113/21.4
m=5.28g


6 0
3 years ago
Read 2 more answers
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
2 years ago
Why do we balance chemical equations
antiseptic1488 [7]
So it could follow the correct mass for the atom
5 0
2 years ago
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