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2 years ago

If the magnitude of charges on these source charges is arranged in a descending order, which is the correct sequence?

Physics

2 answers:

pickupchik [31]2 years ago

8 0

Answer:

Field C > Field A > Field B > Field D

Explanation:

As we know that electric flux due to a point charge is given by the formula

$\phi = \frac{q}{\epsilon_0} = \int E.dA$

so from above equation we can say

If the flux linked with a charge is more than it will have more electric field linked with it.

So from the given four pictures of electric field we can say that maximum electric field is linked with field C as the flux is maximum in the figure

Similarly for field D the flux is minimum so electric field must be least in case D

So we can say more the number of field line then more is the electric field linked with it

bonufazy [111]2 years ago

6 0

The source charges' magnitude is signified by the arrows pointing outward. The more arrows there are, the greater is its magnitude. This is because, each arrow represents an electrical force exerted by the source. When you add up all the arrows there is, the electrical force becomes even greater. The answer in descending order would be C > A > B > D.

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A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo

timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = $\frac{wf}{t}$ = $\frac{1.4 rad/s}{21 s} = 0.067 rad/s2$

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

3 0

2 years ago

A 57 kg paratrooper falls through the air. How much force is pulling him down?

just olya [345]

Answer:

Explanation:

Force = Mass * acceleration due to gravity.

Given

Mass of the paratrooper = 57kg

Acceleration due to gravity = 9.81m/s²

Required

Force pulling him down

Substitute unto formula;

F = 57 * 9.81

Force = 559.17 N

Hence the force pulling him down is 559.17N

7 0

2 years ago

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV

Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

$= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

$= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

8 0

2 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$