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3 years ago

Gravitational potential energy = the energy due to an objects

Physics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

Gravitational energy is the energy of a body when it is taken to a certain height with respect to the earth surface.It is one of the part of potential energy that comes in thr part of the mechanical energy. the work done in taking a body upto a certain height get stored in the body in form of gravtitational potential energy. potential energy is directly proportional to height or altitude upto which the body is taken, mass of the body and gravtational acceleration(9.8m/s^2)So P.E. = mgh

Hope it helps

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Homework help plz it would help a lot

blagie [28]

#82

here we know that

acceleration = 2 m/s/s

time = 5 s

initial speed = 4 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 4 + 2(5)$

$v_f = 14 m/s$

So correct answer will be option D)

#83

here we know that

acceleration = 3 m/s/s

time = 4 s

initial speed = 5 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 5 + 3(4)$

$v_f = 17 m/s$

So correct answer will be option C)

#84

here we know that

acceleration = 7 m/s/s

time = 3 s

initial speed = 8 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 8 + 7(3)$

$v_f = 29 m/s$

So correct answer will be option C)

6 0

3 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

A primary succession does not include any invasive species <br><br> True or false

yan [13]

To be referenced, it would be true

7 0

3 years ago

Rearrange the formula F=ma, and solve for the variable (a)

jeyben [28]

Answer:

a = F/m

Explanation:

So we have to isolate a, in order to do this we need to move m to the other side, and we do that by diving both sides by m, resulting in a = F/m

7 0

3 years ago

A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5

creativ13 [48]

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec

6 0

3 years ago