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3 years ago

A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?

Physics

1 answer:

Lelechka [254]3 years ago

5 0

Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

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The purpose of the fraud fighter/ black light machine is to:

Aleks [24]

The purpose of the fraud fighter or also known as black light machine is that it has the ability to check counterfeited bills. It is designed to be a detector with the counterfeited bills, plus, it is a way of verification. It is composed of a UV light, where in, making it have the ability to detect the counterfeited bills which is its task to do.

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2 years ago

The kinetic theory states that the higher the temperature, the faster the

max2010maxim [7]

Answer: the higher the kinetic energy

Explanation:

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3 years ago

A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi

Marianna [84]

Answer:

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0

3 years ago

A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o

Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

W = P Δ V

W = 1.013 x 10⁵ x (19.7 - 13.7)

W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

now,

change in internal energy

Δ U = Q - W

Q = 3300 x 10³ cal

Q = 3300 x 10³ x 4.186 J

Q = 1.38 x 10⁷ J

now,

Δ U = 1.38 x 10⁷ - 6.029 x 10⁵

Δ U = 1.32 x 10⁷ J

6 0

2 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

3 years ago