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viva [34]
3 years ago
7

A crate slides down a ramp that makes a 20o angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

on it with a 68 N horizontal force. How much work does Paige do on the crate at is slides 3.5 m down the ramp?
Physics
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

-223.64684 J

Explanation:

F = Force that is applied to the crate = 68 N

s = Displacement of the crate = 3.5 m

\theta = Angle between the force and displacement vector = (180-20)

Work done is given by

W=Fscos\theta\\\Rightarrow W=68\times 3.5\times cos(180-20)\\\Rightarrow W=-223.64684\ J

The work that Paige does on the crate is -223.64684 J

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Which element, inductor or capacitor, releases its stored energy before it is removed from the circuit?
Georgia [21]

Inductor releases its stored energy before it is removed from the circuit.

<h3>What is an inductor?</h3>
  • A passive two-terminal electrical device known as an inductor, also known as a coil, choke, or reactor, stores energy in a magnetic field as electric current passes through it.
  • Typically, an inductor is made out of insulated wire twisted into a coil. According to Faraday's law of induction, the time-varying magnetic field generates an electromotive force (e.m.f.) (voltage) in the conductor as the current through the coil varies.
  • Lenz's law states that the induced voltage has an opposing polarity (direction) to the change in current that caused it. Inductors so reject any changes in the current flowing through them.
  • Inductance, which is the ratio of voltage to current change rate, is a property of inductors.

To know more about inductors, refer:

brainly.com/question/4425414

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2 years ago
What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a
andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

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