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3 years ago

What is the kinetic energy of a hammer that starts from rest and decreases its potential energy by 10 kJ?

Physics

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

final kinetic energy of the hammer is 10 kJ

Explanation:

As we know that there is no non conservative force on the system

So here we can use the theory of mechanical energy conservation

So we will have

$\Delta K + \Delta U = 0$

here we know that

$\Delta U = - 10 kJ$

from above expression now

$K_f - K_i - 10 kJ = 0$

$K_f - 0 = 10 kJ$

so final kinetic energy of the hammer is 10 kJ

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When you take pictures with a camera, the distance between the lens and the film or chip has to be adjusted, depending on the di

Shalnov [3]

Answer:

Explanation:

When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).

Therefore, if the picture of an object that is far away, the lens must move towards the film.

The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.

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4 years ago

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston i

AVprozaik [17]

Answer:

$D = 55.2 cm$

Explanation:

As we know that the total mass of the all four players is given as

$M = 4\times 110$

$M = 440 kg$

diameter of the piston of cheer leader is given as

$d_1 = 16 cm$

are of cross-section is given as

$A_1 = \pi r^2$

$A_1 = \pi(0.08)^2 = 0.02 m^2$

mass of the cheer leader is given as

m = 55 kg

so the pressure due to cheer leader is given as

$P_{in} = \frac{mg}{A_1}$

$P_{in} = \frac{55 \times 9.81}{0.02}$

$P_{in} = 26835 Pa$

Now on the other side pressure must be same

so we have

$\frac{Mg}{A} + \rho gH = P_{in}$

$\frac{440 \times 9.8}{A} + (900)(9.8)(1) = 26835$

$A = 0.24 m^2$

$\pi r^2 = 0.24$

$r = 0.276 m$

so diameter on the other side is given as

$D = 2 r$

$D = 55.2 cm$

8 0

3 years ago

For a point charge, how does the potential vary with distance from the point charge, r?

mr_godi [17]

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. $1/r^2$ .

e. $r^2$ .

Answer:

The correct option is C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

$V = \frac{k Q }{r }$

Here we can deduce that the electric potential varies inversely with the distance i.e

$V \ \alpha \ \frac{1}{r}$

3 0

3 years ago

You are asked to design a spring that will give a 1070 kg satellite a speed of 3.75 m/s relative to an orbiting space shuttle. Y

Dvinal [7]

Answer:

$380697.33\ \text{N/m}$

$0.138\ \text{m}$

Explanation:

m = Mass rocket = 1070 kg

v = Velocity of rocket = 3.75 m/s

a = Acceleration of rocket = 5g

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

The energy balance of the system is given by

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow kx=\dfrac{mv^2}{x}\\\Rightarrow kx=\dfrac{1070\times 3.75^2}{x}\\\Rightarrow kx=\dfrac{7250}{x}$

The force balance of the system is given by

$ma=kx\\\Rightarrow m5g=\dfrac{7250}{x}\\\Rightarrow x=\dfrac{7250}{1070\times 5\times 9.81}\\\Rightarrow x=0.138\ \text{m}$

The distance the spring must be compressed is $0.138\ \text{m}$

$k=\dfrac{7250}{x^2}\\\Rightarrow k=\dfrac{7250}{0.138^2}\\\Rightarrow k=380697.33\ \text{N/m}$

The force constant of the spring is $380697.33\ \text{N/m}$ .

4 0

3 years ago

A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago