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3 years ago

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A cat pushes a mouse horizontally on a frictionless floor with a net force of 1.0 N 1.0N1, point, 0, start text, N, end text for

3.0 m 3.0m3, point, 0, start text, m, end text. How much kinetic energy does the mouse gain?

1 answer:

vesna_86 [32]3 years ago

8 0

Answer: 3

Explanation:

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if a current of 0.3 ampere flows through a conductor of resistance of 11.7 ohm's,the voltage across the ends of the conductor is

Gnesinka [82]

Here,

Resistance = 11.7

Ampere = 0.3

Voltage = ?

Now,

R = V/A

11.7 = V/0.3

11.7*0.3 =V

V = 3.51

I hope it's help you...

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7 0

3 years ago

Contrast situations where work is done with different amounts of force to situations where no work is done such as standing stil

uranmaximum [27]

Picking up a sheet of paper . . . work done with small force
Picking up a glass of water . . . work done with moderate force
Picking up a huge boulder . . . work done with a great tremendous force
=================================
Standing still . . .
Holding your tongue out as far as it will go . . .
Holding your arm over your head for 3 days . . .
Holding a huge boulder motionless over your head . . .
Pushing on a brick wall . . .
Pushing as hard as you can against a truck with the wheels locked . . .
. . . . . No work done at all, because the force doesn't move through a distance.

<u>Work done = (force) times (distance)</u>

If the force doesn't move, then the distance is zero, and the work done is zero.

5 0

3 years ago

5) Old-style dimmers consisted of a variable resistor so you could dial in more resistance to lower the current and, hence, dim

NeX [460]

Answer:

The power of the bulb is reduced to a quarter

Explanation:

The power in an electrical circuit is

P = V I = I² R

If the dimmer cuts the current in half

I = I₀ / 2

The resistance of the light bulb after hot is little affected by small changes in the current

We substitute in the equation

P = I₂ R

P = (I₀ /2)² R

P = ¼ I₀² R

P = ¼ P₀

The power of the bulb is reduced to a quarter

4 0

4 years ago

Calculate the mass of an object that is accelerating 14 m/s^2 with a force of 280 N

Masteriza [31]

Answer:

$\boxed {\boxed {\sf 20 \ kilograms }}$

Explanation:

Force is the product of mass and acceleration.

$F=m*a$

We can find mass, since we know the acceleration and force.

The acceleration is 14 meters per square second. The force is 280 Newtons, but we should convert the units to make the problem simpler later on.

1 kilogram meter per square second is equal to 1 Newton.
280 Newtons are equal to 280 kg*m/s²

$F= 280 \ kg*m/s^2 \\a= 14 \ m/s^2$

Substitute the values into the formula.

$280 \ kg*m/s^2 = m* 14 \ m/s^2$

We want to solve for the mass or m. Therefore we must isolate the variable on one side of the equation.

m is being multiplied by 14 m/s². The inverse of multiplication is division. Divide both sides of the equation by 14 m/s²

$\frac{280 \ kg*m/s^2}{14 \ m/s^2}=\frac{m*14 \ m/s^2}{14 \ m/s^2}$

$\frac{280 \ kg*m/s^2}{14 \ m/s^2}=m$

The m/s² will cancel, which is why we converted the units earlier.

$\frac{280 \ kg}{14 }=m$

$20 \ kg=m$

The mass of the object is <u>20 kilograms.</u>

7 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago