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jonny [76]
3 years ago
10

The density of solid copper is 8960 kg/m3.

Physics
1 answer:
Ahat [919]3 years ago
3 0

Explanation:

The density of a solid copper is 8960kg/m^3 while the density of a molten copper is 7900kg/m^3 this is due to the difference in the state of matter, the solid copper is in solid state while the molten copper is in form of a liquid. In the solid copper atoms are tightly pack while they atom are not tightly packed.

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What is 4 differences between saturated unsaturated and supersaturated solutions
Dennis_Churaev [7]

Answer:

Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.

3 0
3 years ago
1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car e
tankabanditka [31]

Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

Acceleration = a = ?

F = ma

a = F/m

a = 2000/ 1000

a = 2 m/s^{2}

Q2: Solution:

Change in velocity = Δv = ?

acceleration = change in velocity / time

a = Δv/t

Δv = axt

Δv = 2 x 10

Δv = 20 m/s

Q3: Solution:

Impulse = I = ?

Impulse = Force x time

I = 2000 x 10

I = 20000 Ns

Q4: Solution:

Change in Momentum = Δp = ?

Δp = mΔv

Δp = 1000 x 20

Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

Δp = m x (vf-vi)

Δp = 1000 x (vf - 0 )

20000 = 1000 x vf

vf = 20000/1000

vf = 20 m/s

Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

Δp = 1000 x (0-20)

Δp = -20000 kgm/s

Q9: Solution:

Impulse = Δp

Impulse = -20000 Ns

Q10: Solution:

Impulse = ?

Impulse = F x t

F = impulse/t

F = -20000/4s

F = -5000 N

Q11: Solution:

F = ma

a = ?

a = F/m

a = -5000/1000

a = -5m/s^{2}

6 0
3 years ago
Which energy changes take place when a pedalling cyclist uses a generator (dynamo) to light his bicycle lamp?
Elina [12.6K]

Answer: chemical → kinetic → electrical → light

Explanation: At the same time when the generator lights up the bicycle lamp the lamp lights up using electrical energy so mechanical energy is also transformed into electrical energy.

8 0
2 years ago
If a person does 20 J of work on a book in 10 seconds, how powerful was the person while doing the work?
Korolek [52]

Answer:

2 W

Explanation:

P = W/t

= 20 J / 10 s

= 2 W

4 0
2 years ago
A water hose is used to fill a large cylindrical storage tank of
ludmilkaskok [199]

Answer:

maximum possible speed by solving above equation for 7D is

v_{max} = \sqrt{\frac{49}{5}gD}

minimum possible value of speed for solving x = 6D is given as

v_{min} = \sqrt{9gD}

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) =  (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

x = (v_o cos45) t

y =  (v_o sin45) t - \frac{1}{2}gt^2

Now from above two equations we have

y = x - \frac{gx^2}{v_o^2}

now we know that height of the cylinder is 2D so we have

x - \frac{gx^2}{v_o^2} = 2D

by solving above equation we have

x = \frac{v_o^2 \pm v_o^2\sqrt{1 - \frac{8gD}{v_o^2}}}{2g}

now we know that maximum value of x is 7D

so the maximum possible speed by solving above equation for 7D is

v_{max} = \sqrt{\frac{49}{5}gD}

minimum possible value of speed for solving x = 6D is given as

v_{min} = \sqrt{9gD}

6 0
3 years ago
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