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zimovet [89]
3 years ago
12

Radiant energy travels in straight lines until it strikes an object where it can be ______,_____or transmitted

Physics
1 answer:
Alborosie3 years ago
7 0
<span>Radiant energy travels in straight lines until it strikes an object where it can be absorbed, reflected or transmitted</span>
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An object is dropped from the top of a building and is observed to take 7. 2s to hit the ground. How tall is the building?.
maw [93]

the height of the building is H=36 m.

<h3>What is The Law of Gravity?</h3>

According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.

Properties of Gravity -

  • It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.
  • It obey inverse square law.
  • It is the weakest force known in nature.

Examples of Gravity -

  • The force that holds the gases in the sun.
  • The force that causes a ball you throw in the air to come down again.
  • The force that causes a car to coast downhill even when you aren't stepping on the gas.

v₀=0 m/s

H₀=0 m

g=10 m/s²

t=7,2 s

H - ?

H= H_{0} + v_{0}t+\frac{gt^{2} }{2}   \\

H = 0 +0 × 7.2 + 10(7.2)²/2

H = 36m

to learn more about Gravity go to - brainly.com/question/12528243

#SPJ4

8 0
1 year ago
2. Which of the following is accurate when discussing specific neat?
MrRissso [65]

Answer:

The specific heat of a gas may be measured at constant pressure. - is accurate when discussing specific heat.

Explanation:

6 0
3 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of the gasoline goes into usefu
Aleksandr [31]

Answer:

686.11 N

1.7733 gallons

Explanation:

\eta = Efficiency = 30%

V = Volume of gasoline

E = Energy content of gasoline = 1.3\times 10^8\ J/gal

F = Force

s = Displacement = 108000 m

v = Velocity

Work done is given by

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{VE\eta}{s}\\\Rightarrow F=\frac{1.9\times 0.3\times 1.3\times 10^8}{108000}\\\Rightarrow F=686.11\ N

The force required to keep the car moving at a constant speed is 686.11 N

Here the force is directly proportional to speed

\\\Rightarrow F=v

\\\Rightarrow \frac{F_1}{v_1}=\frac{F_2}{v_2}\\\Rightarrow F_2=\frac{F_1\times v_2}{v_1}\\\Rightarrow F_2=\frac{686.11\times 28}{30}\\\Rightarrow F_2=640.36\ N

W=F\times s\\\Rightarrow 0.3\times 1.3\times 10^8\times V=640.36\times 108000\\\Rightarrow V=\frac{640.36\times 108000}{0.3\times 1.3\times 10^8}\\\Rightarrow V=1.7733\ gal

The gallons that will be used is 1.7733 gallons

7 0
3 years ago
8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,
Katena32 [7]

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

6 0
2 years ago
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