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olga2289 [7]
3 years ago
11

A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

seconds?
350 W
60025 W
27440 W
1715 W
Physics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

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A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

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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

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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

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3 years ago
Think about lifestyle choices that you make on a daily basis. Choose at least four lifestyle choices which you could change to i
eimsori [14]
A.The four lifestyle choices which I could change are:
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4 0
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Define the term work​
bagirrra123 [75]

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force

Explanation:

at least part of which is applied in the direction of the displacement. ... To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.

5 0
3 years ago
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A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy asso
jasenka [17]

Answer:

U(3)=-43J

Explanation:

Potential energy is minus the integral of Fdx.  Doing the integration yields:

U=\int\limits {6.0x+12}\, dx

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U(0)=20J

so

U(0)=-3(0)^2+12(0)+C=20

C=20

Now for x=3.0m

U(3)=-3*(3)^2-12(3)+20

U(3)=-43J

6 0
3 years ago
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