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makkiz [27]
3 years ago
15

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

Physics
1 answer:
eduard3 years ago
7 0

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.
Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

\frac{Q}{t} = \frac{kA\Delta T}{d}

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, \Delta T, T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}

k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}

\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}

\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}

\Delta T_{air} = 44.9 \Delta T_{glass}

There are two layers of Glass and one layer of Air so the total temperature would be given as,

\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}

\Delta T = 2\Delta T_{glass} +\Delta T_{air}

20\°C = 46.9\Delta T_{glass}

\Delta T_{glass} = 0.426\°C

Finally the rate of heat flow through this windows is given as,

\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}

\Delta {Q}{t} = 0.84*24*10 -3*0.426

\Delta {Q}{t} = 179W

Therefore the correct answer is D. 180W.

3 0
3 years ago
A human services professional has worked with a particular client for a couple of years. The client
zhuklara [117]

Answer:

Respect the client’s decision

Explanation:

just took the test

4 0
3 years ago
1. A 9000 kg van was stopped at a traffic light when it rear-ended with an 850 compact car moving to the east at a velocity of 5
____ [38]

Answer:

1.785 m/s

Explanation:

The momentum can be calculated using the expression below

M1 *V1 + M2 * V2 = (M1+M2) V3

M1= mass of van=9000 kg

M2= mass of car= 850kg

V3= velocity of entangled car

V1= Velocity of the van= 0

V2= velocity of the car= 5 m/ s

Substitute the values

(900×0) + (500×5)=( 900+500)× V3

2500=1400 V3

V3=2500/1400

V3= 1.785 m/s

Hence, velocity of the entangled cars after collision is 1.785 m/s

8 0
2 years ago
Controlling the amount of current in a circuit by opposing the flow of charge
Vera_Pavlovna [14]

Answer:

Resistor

Explanation:

Resistor provides resistance to the flow of electric current in a circuit.

It controls the amount of current in a circuit by acting as a barrier to the flow of electric charges

8 0
3 years ago
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In the gene TATTCATTGTTA—TGATTT—ATTCG, CATTGTTA encodes for pepsin, a digestive enzyme. The rest of the sequence doesn’t code fo
MatroZZZ [7]

Answer:

TATTCATTCATTA—TGATTT—ATTCG

Explanation:

A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.

Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.

Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.

8 0
3 years ago
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