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3 years ago

12

A moving negative charge placed in an external magnetic field circulates counterclockwise in the plane of the paper. In which di

rection is the magnetic field pointing?

1 answer:

docker41 [41]3 years ago

7 0

Answer:out of the page

Explanation:

This can be answered by Lorentz force experienced by a charged particle in a magnetic field

which is given by F=q(vxB)

as the charge is moving in anti-clockwise direction so velocity will be tangential to imaginary circle and the force is acting towards the center so magnetic field is inside the page but the charge is negative so magnetic field is out of page(F is perpendicular to v &B)

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The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

nataly862011 [7]

Answer with Explanation:

Let rest mass $m_0$ at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force= $\frac{GmM}{r^2}$

Distance of P from moon=D-x

$F_m$ =Force applied on rest mass due to m

$F_m$ =Force on rest mass due to mas M

$F_M=F_m$ because net force is equal to 0.

$F_m=F_M$

$\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}$

$\frac{m}{(D-x)^2}=\frac{M}{x^2}$

$\frac{x^2}{(D-x)^2}=\frac{M}{m}$

$\frac{x}{D-x}=\sqrt{\frac{M}{m}}$

Let $R=\sqrt{\frac{M}{m}}$

Then, $\frac{x}{D-x}=R$

$x=DR-xR$

$x+xR=DR$

$x(1+R)=DR$

$x=\frac{DR}{1+R}$

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x= $\frac{2}{3}D$

Net force is zero

$F_m=F_M$

$\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}$

$\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}$

$\frac{M}{m}=4$

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0

4 years ago

Find the displacement of a car that accelerates at -6m/s2 for 5 sec to a final velocity of 60m/s.

Anvisha [2.4K]

Answer:

Explanation:

2as=vf²-vi²

2as=(60)²-0²

2as=3600 m/s

s=3600/2*a

s=3600/2*(-6)

s=3600/-12

s=-300 m

3 0

3 years ago

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake s

WINSTONCH [101]

Answer: a) 5 x 10^5 kg/s b) 444 MW

Explanation:

Kinetic energy per unit mass Ke is

Ke = V^2 / 2

Ke = 3^2 / 2 = 4.5 J/kg = 0.0045 kJ/kg

Now potential energy per unit mass Pe is

Pe = g x z = 9.8 x 90 = 882.9 J/kg = 0.8829 kJ/kg

The total mechanical energy of the River per unit mass e = Ke + Pe = 0.0045 + 0.8829 = 0.88744 J/kg

M = P x V = 1000 x 500 = 5 x 10^5 kg/s

b) power generation potential of the entire river at the location Wmax = Emech = M x Emech = 500,000 x 0.88744 = 444,000kW = 444MW

7 0

4 years ago

Tarzan (mT=85.0 kg) is perched on a cliff 16.0 m above Jane. Jane is in the foreground is looking forwards. In the background, T

poizon [28]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Part A

The velocity of Tarzan is $v_T = 17.7 \ m/s$

Part B

The percentage loss of kinetic energy is %-loss $= 61.4$ %

Part C

The speed would be exactly the same

Explanation:

From the question we are told that

The mass of Tarzan is $m = 85.5 \ kg$

The height of the cliff is $h = 16.0 \ m$

From the law of conservation of energy

$PE = KE$

Where PE is the potential energy of Tarzan at the top of the cliff which is mathematically represented as

$PE = mgh$

KE is the kinetic energy of Tarzan as he swings toward jane and is mathematically represented as

$KE = \frac{1}{2} mv^2_T$

So

$mgh = \frac{1}{2}m v_T^2$

=> $gh = \frac{1}{2} v_T^2$

Substituting values

$9.8 * 16 = \frac{1}{2} v_T^2$

=> $v_T = \sqrt{\frac{9.8 * 16}{0.5} }$

$v_T = 17.7 \ m/s$

From the question we are told that the actual velocity of Tarzan is

$v = 11 m/s$

The kinetic energy at the the velocity is mathematically represented as

$KE_A = \frac{1}{2} m v^2$

Substituting values

$KE_A = \frac{1}{2} (85) * 11^2$

$KE_A = 5142.5 J$

The kinetic energy at $v_T$ is

$KE = \frac{1}{2} * 85 * 17.7^2$

=> $KE = 13314.83\ J$

Now the percentage loss of kinetic energy is mathematically evaluated as

%-loss $= \frac{KE - KE_A}{KE} * 100$

substituting values

%-loss $= \frac{13314.83 - 5142.5}{13314.83} * 100$

%-loss $= 61.4$ %

The mass of Tarzan does not affect the speed because looking at this equation

$gh = \frac{1}{2} v_T^2$

We see that the velocity of Tarzan is not dependent on the mass of Tarzan

4 0

3 years ago

4 kg of water was

morpeh [17]

Given :

Mass =5kg

T

1

=20

∘

C,T

2

=100

∘

C

ΔT=100−20=80

∘

C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

6 0

3 years ago