A stone is thrown straight upward and reaches a maximum height of 31.8 m above its

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

2 years ago

What are 3 things that can affect gravity?

den301095 [7]

The force of gravity the masses exert on each other. If one of the masses is doubled , the force of gravity between the objects is doubled. Increases , the force of gravity decreases.

5 0

2 years ago

Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.

MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>

It is given that, the infinite well having a width of 0.050 mm.
We have the expression for energy of electron in an infinite well as,

$E_n=\frac{n^2h^2}{8mL^2}$

where;

$m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1$

Thus, the lowest energy of electron in an infinite well is,

$E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J$

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

#SPJ4

7 0

10 months ago

Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou

belka [17]

The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year.
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather

4 0

2 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago