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3 years ago

How much power does an Ox pulling a plow 20 m cross a field, exerting 120J of work over a period of 15 s?

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

3 0

Answer:

Explanation:

Given parameters:

Distance covered = 20m

Work done = 120J

Time = 15s

Unknown:

How much power does the Ox exerts = ?

Solution:

Power is the rate at which work is being done

Power = $\frac{Work done }{time}$

Power exerted by ox = $\frac{120}{15}$ = 8W

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In a uranium fission reaction, uranium splits into two smaller atoms and energy. Where did the energy come from?

postnew [5]

Answer: C. Some of uranium's mass is converted into energy, so the smaller atoms have less mass.

Explanation:

From Einstein's mass-energy relation:

E = mc²

Mass and energy are equivalent. Mass can be converted into energy and energy into mass.

When Uranium atoms under go nuclear fission, smaller atoms are formed and huge amount of energy is released. This energy comes from the mass difference of the uranium nuclei and new nuclei formed. This mass converted into energy according to Einstein's equation.

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3 years ago

Read 2 more answers

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

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3 years ago

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

Where:

m: is the proton's mass = 1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

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3 years ago

What are two unique characteristics of a social psychologists?

AleksandrR [38]

Topics examined in social psychology include: the self concept, social cognition, attribution theory, social influence, group processes, prejudice and discrimination, interpersonal processes, aggression, attitudes and stereotypes.

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3 years ago

A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy