A 755 N diver drops from a board 10.0 m above the water's surface. Find

A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of

ANEK [815]

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.

<h3>What is the right-hand thumb rule?</h3>

Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.

The proton will be subject to a magnetic pull that is directed into the page.

Hence, option B is correct.

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4 0

2 years ago

Which of the following statements is true? The square root of the molecular weight of a gas is equal to its rate of diffusion. T

forsale [732]

The diffusion coefficient of the gas is proportional to the average rate of thermal motion of the molecules.

the average velocity is inversely proportional to the square root of the molar mass

so
The gas diffusion rate is inversely proportional to the square root of its molecular weight.

8 0

2 years ago

Read 2 more answers

Which property of gold allows it to be used this way?

densk [106]

the electric conductivity of gold is very high

3 0

3 years ago

MARKING BRAINLIST | Which situation below would have the STRONGEST gravitational force between them?

maks197457 [2]

Case d) has the strongest gravitational force

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

a) For this pair of objects:

m1 = 10 kg

m2 = 2 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N$

b) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N$

c) For this pair of objects:

m1 = 2 kg

m2 = 2 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N$

d) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N$

Therefore, the strongest gravitational force is in case d).

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6 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago