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3 years ago

a. Of the three experiments that are used to confirm the Big Bang theory, which is the most interesting to you and why?

Physics

1 answer:

HACTEHA [7]3 years ago

6 0

Answer:

Explained

Explanation:

Big Bang happened about 13.7 billion years ago. It is most accepted theory of universe creation. Several experiments have confirmed the happening of big bang back then, in promising way. One of this experiment is the red shift experiment. A scientist named Edwin Hubble tried to study the spectrum of stars. He made a shocking revelation that universe is rushing outwards in all directions. Light travelling away registers on the red side. He observed that the distance stars showed red shift in accelerated manner. If the universe is accelerating that means, at one point it would have been smaller, in fact it was as small as an atom, some how it exploded and lead to formation of the expanding universe.

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The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between

Nat2105 [25]

Answer:

14 x 0.27 = 3.78 is your answer

Explanation:

the question is asking for the weight of the object so you multiply and get 3.79

5 0

3 years ago

Are the items in the list renewable or nonrenewable resources?

Elza [17]

N
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8 0

3 years ago

Read 2 more answers

Doing the same amount of work in less time requires more power.<br> O A. True<br> O B. False

padilas [110]

Answer:true

Explanation:

6 0

3 years ago

An airplane has an effective wing surface area of 17.0 m2 that is generating the lift force. In level flight the air speed over

olga_2 [115]

To solve this problem it is necessary to apply the equations given from Bernoulli's principle, which describes the behavior of a liquid moving along a streamline. Mathematically this expression can be given as,

$P_1 + \frac{1}{2}\rho*v_1^2 + P_2 + \frac{1}{2}*\rho*v_2^2=0$

Where,

$P_i =$ Pressure at each state

$\rho$ = Density

$v_i =$ Velocity

Re-organizing the expression we can get that

$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Our values are given as

$v_1 = 40m/s$

$v_2 = 55m/s$

$\rho_{water} = 1.2kg/m^3 \rightarrow$ Normal Conditions

Replacing we have,

$P_1 -P_2 = \frac{1}{2}*1.2*(55^2-40^2)$

$P_1 - P_2 = 855Pa$

If we consider that there is a balance between the two states, the Force provided by gravity is equivalent to the Support Force, therefore

$F_l = F_g$

Here the lift force is the product between the pressure difference previously found by the effective area of the aircraft, while the Force of gravity represents the weight. There,

$F_g = W$

$F_l = (P_2-P_1)A$

Equating,

$(P_1 - P_2)*A = W$

$W = 855*17$

$W = 14535 N$

Therefore the weight of the plane is 14535N

3 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the

ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0

3 years ago