The force needed to move an object 14 meters with 83 joules of work is what?

(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

6. traction a. friction between a tire and the road.b. pushes a moving object out of a curve and into a straight linec. the abil

Gennadij [26K]

Answer:

6. a. friction between a tire and the road

7. c. energy of motion

8. c. the force with which a moving vehicle hits another object

Explanation:

6. As a car moves along the road, the tires push back against the ground. As tires push back against the ground, the road exerts and opposing force to the motion of the tires. This opposing force is the friction between the tires and the road. This opposing force between the tires and the rad is called traction.

So, the answer is a

7. As an object moves, it has energy. This energy due to its motion is called kinetic energy.

So, the answer is c

8. When a moving vehicle hits another object, it exerts a force on the object. The process of the vehicle hitting the other object is called impact and the force exerted on the object is called the force of impact.

So, the answer is c.

4 0

3 years ago

HELLPPPP. The metalloid that has three valence electrons is ___________?

Alja [10]

Answer:

Boron

Explanation:

8 0

3 years ago

If you warm up the volume of a balloon but keep the pressure the same, you would be using which gas law?

SIZIF [17.4K]

Answer:

Charles law

Explanation:

Charle's law states that the volume (V) of a given gas is directly proportional to the absolute temperature (T) at a constant pressure.

That is;

: V ∝ T

: V/T = K

According to this question, the volume of a balloon is warmed up but the pressure is kept the same. Charles law will be used because it shows the relationship between the volume (V) and the temperature (heat) at a constant pressure (P).

3 0

3 years ago