Answer please I need help

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

$v=0+(9.8)(0.75)=7.35 m/s$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

8 0

3 years ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)

6 0

4 years ago

Read 2 more answers

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

3 years ago

Explain how we can deduce the temperature of a star by determining its color.

ValentinkaMS [17]

Answer: Stars are bright and have the ability to emit lights of various wavelength. The color of a star plays a significant role. It helps us in determining its temperature. It ranges from reddish color to a bluish-white color. A red color star indicates that the star is of low temperature, whereas a bluish-white star indicates that the star is of high temperature.

7 0

3 years ago

At what point on the hill will the car have zero gravitational potential energy?

Mila [183]

At the bottom of the hill

8 0

3 years ago

Read 2 more answers