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gulaghasi [49]
3 years ago
5

How is work affected when an object is lifted straight up instead of using a ramp? Work will increase.

Physics
1 answer:
kifflom [539]3 years ago
3 0
Without friction, the amount of work only depends on the final height,
and is not affected by the route used to get there. 

If the ramp has no friction, then it has no effect on the total amount
of work done.  The work to lift the load straight up is the same.

If the ramp has some friction, then it takes more work to use the ramp
than to lift the load straight up.  Then the work to lift the load straight up
would be less than when the ramp is used.


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devlian [24]

Answer:R=1.424 M\Omega

Explanation:

Given

mass of gold m=1.6 gm

Length of wire L=2.2 km

Resistivity of gold \rho =2.44\times 10^{-8}

density of gold =19.3\times 10^3 kg/m^3

and mass=volume\times density

1.6\times 10^{-3}=volume\times 19.3\times 10^3

volume=8.29\times 10^{-8} m^3

And Resistance R=\frac{\rho L}{A}

also be written as

R=\frac{\rho L^2}{V}

where L=length

V=volume

\rho =resistivity\ of\ gold

R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}

R=1.42\times 10^{6} \Omega

R=1.424 M\Omega

5 0
3 years ago
The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the
s344n2d4d5 [400]

Answer:

j

Explanation:

x = 4 t^2 - 2 t - 4.5

Position at t = 3 s

x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m

Velocity at t = 3 s

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v ( t = 3 s) = 8 x 3 - 2 = 22 m/s

Acceleration at t = 3 s

a = dv / dt = 8

a ( t = 3 s ) = 8 m/s^2

When is the velocity = 0

v = 0

8 t - 2 = 0

t = 0.25 second

When is the position = 0

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t = \frac{2 \pm \sqrt{4 + 72}}{8}

t = 1.4 second

8 0
3 years ago
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The gas and food are sources of energy. They act as fuel. This energy is utilized to perform work. The gas is used to run the car. The food is metabolized inside the body which is a source of energy and utilized to perform every day work.

Hence, The gas and food are examples of energy.

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