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3 years ago

An arrow is used to show a force acting on an object. what information does the arrow give about the force

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

6 0

The Force surrounds us and penetrates us. It flows through Jedis. May the Force be with you all.

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Which of these orders has the structure of the universe from the smallest to largest?

Oliga [24]

Answer:

From largest to smallest they are: Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:

6 0

2 years ago

What is the gravitational potential energy of a two-particle system with masses 4.5 kg and 6.3 kg, if they are separated by 1.7

zepelin [54]

Answer:

7.41 x 10^-10 J

Explanation:

m = 4.5 kg

M = 6.3 kg

d = 1.7 m

The formula for the gravitational potential energy is given by

$U = \frac{-GMm}{d}$

Where, G is the Universal gravitational constant

G = 6.67 x 10^-11 Nm^2/kg^2

$U = \frac{-6.67 \times 10^{-11}\times 6.3 \times 4.5}{1.7}$

U = - 1.112 x 10^-9 J

Now the separation is tripled, d' = 3 x d = 3 x 1.7 m = 5.1 m

Let the potential energy is U'

The formula for the gravitational potential energy is given by

$U' = \frac{-GMm}{d'}$

Where, G is the Universal gravitational constant

G = 6.67 x 10^-11 Nm^2/kg^2

$U' = \frac{-6.67 \times 10^{-11}\times 6.3 \times 4.5}{5.1}$

U' = - 3.71 x 10^-10 J

the work done is equal to the change in potential energy

W = U' - U

W = - 3.71 x 10^-10 + 1.112 x 10^-9

W = 7.41 x 10^-10 J

7 0

3 years ago

A train is travelling at 15m/s. It accelerates at 3m/s² for 20 seconds. How far does the train travel in this time?

Fittoniya [83]

Answer:

Explanation:

The answer is 900m

4 0

3 years ago

Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti

Triss [41]

Answer:

<em>The total potential (magnitude only) is 11045.45 V</em>

Explanation:

<u>Electric Potential </u>

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

$\displaystyle V=\frac{kq}{r}$

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

$\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V$

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

$V_2=-22090.91\ V$

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

$V_3=+22090.91\ V$

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

$\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V$

The total electric potential in A is

$V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V$

$V=-11045.45 \ V$

The total potential (magnitude only) is 11045.45 V

8 0

2 years ago

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

$a=cs$ (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

$v^2=v_o^2+2a \Delta s$ (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

$v^2=2(cs)\Delta s$

You do the constant c in the last equation, then you replace the values of v, s and Δs:

$c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44$

6 0

3 years ago