The force of gravity is less between two objects that

a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am

Sindrei [870]

Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

θ = angle between wire's current direction and magnetic field = 90°

0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

3 0

3 years ago

A sample of oxygen gas at 25.0Â°c has its pressure tripled while its volume is halved. What is the final temperature of the gas?

Marat540 [252]

Answer:

447 K

Explanation:

25 C = 25 + 273 = 298 K

Assuming ideal gas, we can apply the ideal gas law

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}$

Since pressure is tripled, then $P_2 / P_1 = 3$ . Volume is halved, then $V_2 / V_1 = 0.5$

$T_2 = 298*3*0.5 = 447 K$

4 0

3 years ago

Read 2 more answers

A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?

amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

3 0

3 years ago

Read 2 more answers

A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bu

Mariulka [41]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

3 0

2 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

2 years ago