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marusya05 [52]
3 years ago
11

Please help urgent i need help easy queston

Physics
1 answer:
mihalych1998 [28]3 years ago
3 0
I believe it’s B but i’m not sure
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A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
3 years ago
What is the potential energy of a 2500 g object suspended 5 m above the earth's surface?
alexandr402 [8]

Answer:

Explanation:

Potential energy, which is the energy a body assumes at a position, can be calculated using the formula:

P.E = m × g × h

Where;

m = mass (kg)

g = acceleration due to gravity (10m/s²)

h = height (m)

7 0
3 years ago
When all else remains the same, what effect would decreasing the focal length have on a convex lens?
aniked [119]
<h3>Answer;</h3>

<u>It would make the lens stronger. </u>

<h3>Explanation;</h3>
  • The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
  • The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
  • Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
  • Thus; decreasing the focal length of a convex lens makes the lens stronger.

4 0
3 years ago
Read 2 more answers
Questlon 2 of 10
pochemuha

Answer:

the answer is OD which says tge buoyant girce on an object is equal to the weight of th fluid

3 0
3 years ago
A car's bumper is designed to withstand a 6.84 km/h (1.9-m/s) collision with an immovable object without damage to the body of t
den301095 [7]

Answer:

F = 5.256 x 10^{3} N

Explanation:

From the work energy theorem we know that:

The net work done on a particle equals the change in the particles kinetic energy:

W = F.d, ΔK =\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}  , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}

where:

W = work done by the force

F = Force

d = Distance travelled

m = Mass of the car

vf, vi = final and initial velocity of the car

kf, ki = final and initial kinetic energy of the car

Given the parameters;

m = 830kg

vi = 1.9 m/s

vf = 0 km/h

d = 0.285 m

Inserting the information we have:

F.d = \frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}

F = \frac{\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i} }{d}

F = \frac{ 0   - \frac{1}{2}  X830 X 1.9^{2} }{0.285}

F = 5.256 x 10^{3} N

3 0
3 years ago
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