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Monica [59]
1 year ago
13

B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find

the force required at the plunger.
a. 675.2kgf

b. 67.52kgf

c. 6752.0kgf

d. None of the above
Physics
1 answer:
lozanna [386]1 year ago
4 0

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

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If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.
Alexus [3.1K]

Answer:

θ₂ = 90° - θ₁

Explanation:

When the light falls on a mirror it bounces back. This is know as reflection. The incident angle is equal to the angle of reflection.

Here, the light strikes the mirror at an angle = θ₁

To find the angle of reflection we first need to understand angle of incidence. The angle of incidence is the angle made between the incident ray and normal. Normal is an imaginary line drawn perpendicular line on the boundary of the mirror.

Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

Angle of incidence = 90° - θ₁.

Thus, angle of reflection, θ₂ = 90° - θ₁

3 0
2 years ago
An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The cur
MariettaO [177]

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H

is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

\Delta T=100-21\ ^0C=79\ ^0C

So,  

\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ

Heat Supplied Q=VIt

where i=current

V=Voltage

8.5\times 120\times t=2.4\times 4186\times 79

t=778.10 s

 

6 0
2 years ago
A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.
Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0
2 years ago
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