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3 years ago

Which two formulas are used to calculate potential and kinetic energy

Physics

2 answers:

goblinko [34]3 years ago

6 0

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

kogti [31]3 years ago

3 0

Answer:

$\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}$

Explanation:

Kinetic energy formula

$\displaystyle KE=\frac{1}{2} mv^2$

Potential energy formula

$\displaystyle PE=mgh$

$\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)$

$\displaystyle PE \Rightarrow \sf potential \ energy \ (J)$

$\displaystyle m \Rightarrow \sf mass \ (kg)$

$\displaystyle v \Rightarrow \sf velocity \ (m/s)$

$\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)$

$\displaystyle h \Rightarrow \sf height \ (m)$

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WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME

koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} +(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

4 0

3 years ago

A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time t to fall a vertical distance y and at the same time travels a horizontal distance x.

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance y using the equation,

$y=u_yt+\frac{1}{2} gt^2$

Substitute 0 m/s for $u_y$ , 9.81 m/s²for g and 1.5 m for y.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

Substitute 500 m/s for $u_x$ and 0.5530s for t.

$x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m$

The bullet travels a distance of 276.5 m.

5 0

3 years ago

Part of understanding the physical effects on Mars, we must understand

Ber [7]

Answer:

so easy add the subtract then multiplay the add

Explanation:

8 0

3 years ago

A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe

Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

= (1410 - 207)/(80.5/60)

= 60(1410 - 207)/80.5

= 60(1203)80.5

= 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

= 207 × 1.342 + 1/2 × 896.65 × 1.342²

= 277.725 + 807.417

= 1085.14 revolutions ≅ 1085 revolutions

5 0

3 years ago

What is 8 m/s - 3 m/s?

Dima020 [189]

5m/s 8-3 is five so five

3 0

3 years ago