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olga nikolaevna [1]
3 years ago
7

HURRY NEED HELP ASAP

Physics
1 answer:
gavmur [86]3 years ago
8 0

Explanation:

The electrostatic force is given by :

F_E=k\dfrac{q_1q_2}{r^2} ....(1)

k is electric constant

r is the distance between charges q₁ and q₂

The gravitational force between two masses m₁ and m₂ is given by :

F_G=\dfrac{Gm_1m_2}{r^2} ...(2)

G is gravitational constant

Dividing equation (2) by (1) we get :

\dfrac{F_G}{F_E}=\dfrac{\dfrac{Gm_1m_2}{r^2}}{\dfrac{kq_1q_2}{r^2}}\\\\\dfrac{F_G}{F_E}=\dfrac{Gm_1m_2}{kq_1q_2}

Option (A) is correct.

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Does this curved graph show a function explain how you know
pochemuha

Answer:

No; the graph fails the vertical line test.

The vertical line test is a tool used to determine if we have a function. If we can draw a single straight vertical line through more than one point on the red curve, then the graph is said to have failed the vertical line test. Consequently, this leads to the relation not being a function.

For this circle graph, we can draw a vertical line through more than one point, which is why we don't have a function here.

Put another way, there are inputs (x) that produce more than one output (y), so that's why we don't have a function.

-BBBM

8 0
3 years ago
A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

6 0
3 years ago
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

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3 years ago
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