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3 years ago

How can magnetic and electric fields be demonstrated?

Physics

2 answers:

liubo4ka [24]3 years ago

6 0

Answer:

To demonstrate that an electric current (i.e., moving electric charge) generates a magnetic field, all you need do is simply place a magnetic compass next to a wire in a circuit. When current is passed through the wire, the compass will deflect, indicating the presence of a magnetic field circling the wire.

Explanation:

shepuryov [24]3 years ago

4 0

Electric field is always associated with magnetic field.

Explanation:

Relationship between electric and magnetic fields can be explained by the following example. Consider a wire and a magnetic compass placed close together.

When current is flowing through the wire, the needle in the magnetic compass will deflect showing some value indicating the presence of magnetic field during the experiment. This indicates that the electric field is generally accompanied with the magnetic field.

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A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring,

Mademuasel [1]

Answer:

Explanation:

Given

mass of cannon and cart is $m_1=6475.91 kg$

Spring constant $k=20000 N/m$

mass of Projectile $m_2=234 kg$

Launch Velocity of cannon $v_2=170 m/s$

Launch angle $\theta =31.3^{\circ}$

As the External Force is zero therefore we can conserve momentum

Initially both Projectile and cannon is at rest

Conserving momentum in horizontal direction

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\cos 31.3$

$0+0=6475.91\times v_1+234\times 170\cdot \cos (31.3)$

$v_1=-\frac{234\times 170\cdot \cos (31.3)}{6475.91}$

$v_1=-5.24 m/s$

i.e. cannon is moving in opposite direction of Projectile

6 0

4 years ago

A car travels a distance of 140 km at 70.0 km/hr. It then travels an additional distance of 60.0 km at 40.0 km/hr. The average s

Natali5045456 [20]

Answer:

57.1 km/hr

Explanation:

To find the average speed you take the total distance divided by the total elapsed time.

So, the total distance is 140 + 60 = 200

the total elapsed time is found by taking 140/70=2 and 60/40=1.5

2+1.5=3.5

The plug the numbers into the equation,

200/3.5=57.1

5 0

3 years ago

A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp

KatRina [158]

Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂ ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

Where;

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁] ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂ -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T = 21848.2 / 1680

=> 35 - T = 13

=> T = 35 - 13

=> T = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat ( $Q_{W}$ ) due to the hot water will be equal to the negative of the one ( $Q_{I}$ ) due to the melted ice.

i.e

$Q_{W}$ = - $Q_{I}$ -----------------(a)

Where;

$Q_{I}$ = $m_{I}$ x $c_{I}$ x Δ $T_{I}$ [ $m_{I}$ = mass of ice, $c_{I}$ = specific heat capacity of melted ice which is now water and Δ $T_{I}$ = change in temperature of the melted ice]

and

$Q_{W}$ = $m_{W}$ x $c_{W}$ x Δ $T_{W}$

[ $m_{W}$ = mass of water, $c_{W}$ = specific heat capacity of water and Δ $T_{W}$ = change in temperature of the water]

Substitute the values of $Q_{W}$ and $Q_{I}$ into equation (a) as follows

$m_{W}$ x $c_{W}$ x Δ $T_{W}$ = - $m_{I}$ x $c_{I}$ x Δ $T_{I}$

Note that $c_{W}$ and $c_{I}$ are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$ -----------------------(b)

Now, let's analyse Δ $T_{W}$ and Δ $T_{I}$ . The final temperature ( $T_{F}$ ) of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ $T_{W}$ = change in temperature of water = final temperature of water( $T_{F}$ ) - initial temperature of water( $T_{IW}$ )

Δ $T_{W}$ = $T_{F}$ - $T_{IW}$

Where;

$T_{IW}$ = 22°C [which is the final temperature of water before mixture]

=> Δ $T_{I}$ = change in temperature of melted ice = final temperature of water( $T_{F}$ ) - initial temperature of melted ice ( $T_{II}$ )

Δ $T_{I}$ = $T_{F}$ - $T_{II}$

$T_{II}$ = 0°C (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$

0.400 x ( $T_{F}$ - $T_{IW}$ ) = -0.0550 x ( $T_{F}$ - $T_{II}$ )

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ - 0)

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ )

0.400 $T_{F}$ - 8.8 = -0.0550 $T_{F}$

0.400 $T_{F}$ + 0.0550 $T_{F}$ = 8.8

0.455 $T_{F}$ = 8.8

$T_{F}$ = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0

3 years ago

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

Data:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

3 0

3 years ago

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul

Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

$\alpha = \frac{\omega_{f}- \omega_{i}}{t}$ Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity (rad/s)

ωi : initial angular velocity (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

$\alpha = \frac{20 -0}{0.65}$

α = 30.77 rad/s²

Newton's second law in rotation:

F = ma has the equivalent for rotation:

τ = I * α Formula (2)

where:

τ : It is the net torque applied to the body. (N*m)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the net torque acting on the CD

Data

α = 30.77 rad/s²

I = 3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

3 0

3 years ago