+ 9. What force (in units of Newtons) is needed to give a 3.5 m/s" acceleration to a 1200 kg car?

George Of The Jungle's wife, Mrs. Of The Jungle, has been pestering him to go on a diet. He should have listened. During his com

Elena L [17]

Answer:

112.06 kg - Thats heavy !

Explanation:

Let's do force balance here. Let the object of our interest be George. The forces acting on him are the tension in the upward direction, his weight in the downward direction and the centrifugal force in the downward direction. Considering the upward and downward directions on the y-axis and f=given the fact that George doesn't move up or down, the forces are balanced along the y-axis. Hence doing force balance:

magnitude of forces upward =magnitude of forces downward

i.e., Tension(T) = Weight(mg) + Centrifugal force (mv²/r)

where: 'm' is the mass of George, g is the acceleration due to gravity (9.8 m/s²). v is the speed with which George moves (14.1 m/s) and r is the radius of the circle in which he's moving at the instant (Here since he's swinging on the rope, he moves in a circle with radius as the length of the rope and hence r=7.3m).

therefore, T = m (9.8 + (14.1)²/7.3) = 4150 N

Therefore, m = 112.06 kg

7 0

3 years ago

What do astronomers expect will eventually happen to the Milky Way galaxy?

Nataly [62]

<span> The Andromeda Milkyway is a galactic collision predicted to occur in 4 billion years. I believe thats the answer</span>

5 0

3 years ago

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A gas hot water heater is 65% efficient. The energy output to heat the water is 2.3 kWh.

Art [367]

Answer:

3.5 kWh

1.2 kWh are lost

Explanation:

65=2.3/Ein*100

0.65=2.3/Ein

Ein=2.3/.65

Ein=3.5

3.5-2.3=1.2 kWh

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

the mass of a proton is 1.6726 * 10^-24 and the mass of a electron is 9.1093 * 10^-28. What is the difference between the mass o

crimeas [40]

Proton : 0.00000000000000000000000161726

Electron : 0.00000000000000000000000000091093

Take them away =
1.616349e−24
(that's what it gave me on the calculator when I did proton - electron)

3 0

3 years ago

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