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2 years ago

1. What coefficients would balance the following equation?

Physics

2 answers:

Dovator [93]2 years ago

4 0

Answer:

Explanation:

Stolb23 [73]2 years ago

3 0

Answer:

no question matee

Explanation:

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The amount of gravitational potential energy released as:_________.

nordsb [41]

Answer:

b.it depends on the distance it falls

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2 years ago

A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th

vivado [14]

Answer:

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

$\dfrac{20}{30}\times 30=20\ cm$

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

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2 years ago

Science is the process of pursuing knowledge?

tatyana61 [14]

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3 years ago

Read 2 more answers

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

2 years ago

When sitting in the tree the cat has a total of 1375J in its Gravitational potential store. What is the maximum amount of energy

Murrr4er [49]

Answer:

1375J

Explanation:

The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.

P.E = mgh

mass = m

Acceleration due to gravity = g

height = h

At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.

Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.

4 0

2 years ago