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Xelga [282]
3 years ago
6

14. measuring cylinder containing some water stands on a scale pan. A solid ball is lowered into the water. The water level rise

s from the 30cm mark to the 40cm' mark. The scale reading increases from 100g to 180. What is the density of the material of the ball?​
Physics
1 answer:
Basile [38]3 years ago
8 0

Answer:

Let d be the density of fluid.

So , Initial reading of balance, F1 =30dg N

After the level reaches 50cm^3

Final reading of balance , F2 =50dg N

Given that difference between final and initial reading is 30g

i.e, F2 −F1

=30 g

⟹50dg−30dg=30g

⟹20dg=30g

⟹d=30g/20g

⟹d=1.5g/cm^3

So, density of fluid is 1.5g/cm^3

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The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?
AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0
3 years ago
The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.
pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0
3 years ago
If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di
masya89 [10]
The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5
8 0
3 years ago
Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third
seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒   F₄ = - (F₁ + F₂ + F₃)

⇒   F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0
3 years ago
2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.
Sergio039 [100]

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

8 0
3 years ago
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