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lisov135 [29]
2 years ago
5

What happens to the change in the value of the speed as you increased the amount of force applied on your

Physics
1 answer:
nataly862011 [7]2 years ago
3 0
The change of speed is proportional to the Force. If you double the Force, you double the change of speed.
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Which two forces operate over the longest distances?
Harman [31]

Answer:

D Electromagnetic and gravitational

5 0
3 years ago
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m]
Tanzania [10]

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

6 0
3 years ago
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
mash [69]

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

250 m = (0 m/s) (17 s) + ½ a (17 s)²

a = 1.73 m/s²

5 0
2 years ago
If 5000Pa a pressure is exerted on an object with the contact area 0.04m^2. Calculate the mass of an object
Whitepunk [10]
  • Pressure = 5000 Pa
  • Contact Area = 0.04 m^2
  • Acceleration due to gravity = 9.8 m/s^2
  • Let the force be F.
  • We know, Force = Pressure × Contact Area
  • Therefore, Force = 5000 Pa × 0.04 m/s^2
  • or, Force = 200 N
  • We know, force = mass × acceleration
  • Therefore, mass = force ÷ acceleration
  • or, mass = 200 N ÷ 9.8 m/s^2 = 20.4 Kg

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>.</u><u>4</u><u> </u><u>Kg</u>

Hope you could understand.

If you have any query, feel free to ask.

6 0
2 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
3 years ago
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