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Anon25 [30]
2 years ago
7

A current-carrying wire 1.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10 A and there is

a resultant force of 3 N on the wire due to the interaction of the current and field, what is the magnetic field strength
Physics
1 answer:
LenaWriter [7]2 years ago
8 0

Answer:

0.2

Explanation:

F= BIL sin©

3= B×10×1.5 sin90

B=3/15

B= 0.2

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When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in
HACTEHA [7]

Answer: -3 meters/second^2.

Explanation:

3 0
3 years ago
Student A states that when she sits down on a chair, she is exerting a force on the chair and that is all that happens. Student
zvonat [6]

Answer:

C

Explanation:

I got it right on the test !!

7 0
3 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
2 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
2 years ago
How long does it take for a 3.5 kW electric water heater to heat 40 kg of water? from 20 ° C to 75 ° C? The specific heat capaci
iren [92.7K]

Answer:

2633.7 s

Explanation:

From the question,

Heat lost by the water heater = Heat gained by the water

Applying,

P = cm(t₂-t₁)/t.................. Equation 1

Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time

make t the subject of the equation

t = cm(t₂-t₁)/P.............. Equation 2

From the question,

Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C

Substitute these values into equation 2

t = 4190×40(75-20)/3500

t = 9218000/3500

t = 2633.7 s

4 0
2 years ago
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