Status

How many of each component are shown in the diagram? Check all that apply.

g100num [7]

Can you show the diagram please

3 0

3 years ago

What happens to the temperature of a substance when it is changing state of matter?

noname [10]

Solid substances have molecules held tightly and close together
Liquid substances have molecules moving loosely
Gaseous molecules are moving completely freely

As moleclues get further apart, i.e. As a substance changes state from solid to liquid to gas, molecules gain kinetic energy and vibrate/move more. This means they gain heat energy (the averge energy a substance has) so the temperature increases

Substances exist in different states at different temperatures and different substances will exist in different states at the same temperature. This is to do with the forces between molecules and how much heat (energy) is required to break them

5 0

3 years ago

20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball

Dovator [93]

The mechanical energy for the first and the second ball is

$10 ^{7} \: joules.$

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

$= \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}$

$= \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}$

$= 18478.69 \: m$

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

$= \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}$

$= \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m$

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

$= 20 \times 9.8 \times 18478.64 \times \frac{20}{2} (1000 \: cos37 °)$

$= 10 ^{7} The potential energy at the maximum height, = m _{2}gh$

$= 20 \times 9.8 \times 51020.41$

$= 10 ^{7} \:J$

Therefore, the total mechanical energy for the first and the

$\:second \: cannonball \: is \: 10 ^{7} \:joules.$

To know about energy, refer to the below link:

brainly.com/question/1932868

#SPJ4

5 0

2 years ago

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

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

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers