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s2008m [1.1K]
3 years ago
10

Conversion fraction 1$=4q, how many are in 20$

Physics
2 answers:
Tom [10]3 years ago
8 0

Answer:

\boxed{\sf 20 \$ = 80q}

Given:

1$ = 4q

To Find:

How many quarters are in 20$

Explanation:

To find out how many quarters are in 20$ we need to multiple 4 × 20.

\sf 1\$  = 4q

\sf  20\$  = 4 \times 20q

\sf = 80q

Over [174]3 years ago
7 0

Answer:

\boxed{\sf 20 \$ = 80q}

Given:

1$ = 4q

To Find:

How many quarters are in 20$

Explanation:

To find out how many quarters are in 20$ we need to multiple 4 × 20.

\sf 1\$  = 4q

\sf  20\$  = 4 \times 20q

\sf = 80q

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Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from re
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Answer:

Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).

Explanation:

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3 years ago
What is newton's second law of motion?
lord [1]

Answer:

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3 0
3 years ago
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca
Mariulka [41]

i believe it is C....tell me if im right please<3

4 0
3 years ago
Read 3 more answers
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
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