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Yanka [14]
3 years ago
8

The type of image formed by a concave mirror depends on whether a reflected object is located ________________ the focal point a

nd mirror or beyond the focal point.
Please fill in the blank space with the correct answer!!!1
Physics
1 answer:
Inga [223]3 years ago
5 0
The answer to fill in the blank is between.
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Why does changing the shape of an object have no effect on the density?
iren [92.7K]
Because the object is still made of the same material 
Density is not affected by the weight and shape of an object its affected by how concentrated the atoms are in a given volume 
7 0
3 years ago
8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x
Alja [10]

Answer:

a.1017.9N

b.1029.7N

c.T=380.6N

d.11.8 N

Explanation:

(a) The absolute pressure at the level of the top of the block is

Po=atmospheric pressure

g=gravity

h1=height

rh0=density of water

P1=Po+rho*g*h1

1.013*10^5+1000(9.81)(0.05)

1.0179*10^5Pa

at the level of the bottom of the block we have

P2=Po+rho*g*h2

1.013*10^5+1000(9.81)(0.17)

1.0297* 10^5 Pa

the downward force exerted on the top by the water is

Ftop=P*A== × = 1.0179* 10^5* 0.100

= 1017.9 N

and the upward force the water exerts on the bottom of the block , which is the buoyant force

Fbot== × = 1.0297 10^5 * 0.100 m

= 1029.7 N

(b) The scale reading is the tension, T, in the cord supporting the block.

if the block is at equilibrium, then sum of vertical forces

EFy=T+Fbot-Ftop-mg=0

T=mg+Ftop-Fbot

T=40*9.81-(1029.7-1017.9)

T=380.6N

(c)  Archimedes principle state that, the buoyant force on the block equals the weight of the displaced water. Thus,

Buoyant force=rho *g*h

= = 1000* 0.100^2 * 0.120 m *9.80 m s=11.8 N

from the answer a Ftop-Fbot

1029.7-1017.9=11.8N, the same as the buoyant force

5 0
3 years ago
Why is it impossible to build a machine that produces more energy than it uses?
Irina18 [472]
Because then it could mess up the machine with to much energy
8 0
3 years ago
A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

\rho = \frac{m}{V} \rightarrow m = V\rho

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

7 0
3 years ago
11. Which of the following is a proper unit of
Sergeeva-Olga [200]
D is the correct answer!!
4 0
2 years ago
Read 2 more answers
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