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Rus_ich [418]
2 years ago
12

On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on

your face, he says that you'll experience a greater buoyant force if you fall overboard than your friends who wear regular-sized Styrofoam-filled life preservers. True or False
Physics
1 answer:
Lelu [443]2 years ago
8 0

Answer:

True.

Explanation:

He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater.  Hence you will be drown inside the water.  

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Identify Label these cell parts
Kruka [31]

Answer:

I don't know this answer because I am not American

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2 years ago
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Stars give off a tremendous amount of thermal and ________ energy.(1 point) chemical chemical electrical electrical nuclear nucl
Olegator [25]
I believe the answer is Nuclear energy
8 0
1 year ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni
Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A

That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

7 0
3 years ago
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is  12 V

Answer:

The value  is  N_s  =  21 \  turns

Explanation:

From the equation we are told that

   The input voltage is  V_{in}  = 120 \ V

   The number of turns of the primary coil is N_p =  210 \  turn

    The output from the secondary is V_o =  12V

From the transformer equation

   \frac{N_p}{V_{in}}  =\frac{N_s}{V_o}

Here N_s is the number of turns in the secondary coil

=> N_s  =  \frac{N_p}{V_{in}}  *  V_s

=>N_s  =  \frac{210}{120}  *  12

=>N_s  =  21 \  turns

4 0
3 years ago
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