PHYSICS COLLEGE

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west.A)If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.

Use 1.602×10−19 C for the magnitude of the charge on an electron.

B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

a-from north to south

b-from south to north

Answers

Answer 1

Answer:

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

$v_(f)^(2)=v_(0)^(2)+2gy\nv_(f)=\sqrt{0+2(9.8(m)/(s^(2)))(145m)}=53.31(m)/(s)$

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

$F=(3.8*10^(8))(1.602*10^(-19)C)(53.31(m)/(s))(0.205T)=6.65*10^(-10)N$

b. south - north (by the rigth hand rule)

I hope this is usefull for you

regards

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What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

$P=68*6894.74\ Pa$

$P=468842.32\ Pa$

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

$P=(1)/(2)\rho v^2$

$468842.32=0.5*1000* v^2$

$v=\sqrt{(468842.32)/(0.5*1000)}$

$v=30.62\ m/s$

Now, We will convert m/s to ft/s

$v =30.62*3.281$

$v=100.46\ ft/s$

Hence, The velocity at discharge is 100.46 ft/s

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

Learn more about Pressure and Velocity here:

brainly.com/question/32124707

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The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.How many capillaries are there?

Answers

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

$A_1v_1=nA_2v_2$ (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

$A=\pi r^2\n\nA_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\n\nA_2=\pi r_2^2=\pi (5*10^(-4)cm)^2=7.85*10^(-7)cm^2$

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

$n=(A_1v_1)/(A_2v_2)=((3.1415cm^2)(40cm/s))/((7.85*10^(-7)cm^2)(0.10cm/s))=1.6*10^9$

Then, the number of capillaries is 1.6*10^9

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:a) the midpoint between the two rings?b) the center of the left ring?

Answers

The electric field strength at the midpoint of the rings is 0 N/C.

The electric field strength at the center of the left ring is 2710.84 N/C.

The given parameters:

Diameter of the rings, d = 10 cm
Distance between the rings, r = 25 cm
Charge of the rings, q = 20 nC

The electric field strength at the midpoint of the rings is calculated as follows;

$E_(net) = E_1 + E_2\n\nE_(net) = E_1(+ve) - E_2(-ve) = 0$

The electric field strength at the center of the left ring is calculated as follows;

$E = (kqL)/((R^2 + L^2)^(3/2)) \n\nE = (9* 10^9 * 20* 10^(-9) * 0.25 \ )/((0.05^2 + 0.25^2 )^(3/2)) \n\nE = 2710.84 \ N/C$

Learn more about electric field here: brainly.com/question/14372859

Final answer:

The electric field strength at the midpoint between the two rings is zero, and at the center of the left ring, it is 2.88 * 10^4 N/C.

Explanation:

The electric field strength at the

midpoint between the two rings is zero. The electric fields from each ring cancel each other out at this point because they are equal in magnitude and opposite in direction.
At the center of the left ring, the electric field strength can be calculated using the formula for the electric field due to a charged ring. The formula is E = k * (Q / r²), where E is the electric field strength, k is the Coulomb's constant, Q is the charge, and r is the distance from the center of the ring. Plugging in the values, we get:

E = (8.99 * 10^9 Nm²/C²) * (20.0 * 10^-9 C) / (0.05 m)² = 2.88 * 10^4 N/C

If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

Answers

pressure absolute = pressure gage + pressure atmosphere

Answer:

650.280

Explanation: 100kpa + 550.280kpa

Two long ideal solenoids (with radii 20 mm and 30 mm, respectively) have the same number of turns of wire per unit length. The solenoid is mounted inside the larger, along a common axis. The magnetic field with in the inner solenoid is zero. The current in the inner solenoid must be: a. two-thirds the current in the outer solenoid
b. one-third the current in the outer solenoid
c. twice the current in the outer solenoid
d. half of the current in the outer solenoid
e. the same as the current in the outer solenoid

Answers

Answer: The current in the inner solenoid is the same as the current in the outer solenoid.

The correct option is e

Explanation: Please see the attachment below

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Answers

Answer: Skinfold testing

Explanation:

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It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

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