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aleksandr82 [10.1K]
3 years ago
15

The diagram below illustrates the law of reflection.

Physics
1 answer:
Luba_88 [7]3 years ago
6 0
The measure of angle B equals the measure of angle C
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Convert 141.2 kg to lbs and show all units and work​
zavuch27 [327]
Your answer is 311.29271 lbs
5 0
3 years ago
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
2 years ago
When a object is at rest, what is it’s speed??
nadya68 [22]

Answer:

0   (there is no speed)

Explanation:

If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.

3 0
3 years ago
Read 2 more answers
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r
andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature T_1=781 K

Reservoir 2 Temperature T_2=335 K

Let Q is the amount of heat Flows i.e. Q=1477 J

thus change in Entropy is given by \frac{\sum Q}{T}

\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}

\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}

\Delta S=\frac{\sum Q}{T}=-1.891+4.4089

\Delta S=\frac{\sum Q}{T}=2.517 J/K                              

6 0
3 years ago
Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

T_H  = 0.5T_L

now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

η_{th = 1 - T_L/0.5T_L

η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

Q_H = 120 kJ

Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

60 = 120 -  Q_L

Q_L = 60 kJ

now, the amount of heat rejection per kg of steam is;

q_L = Q_L/m

we substitute

q_L = 60/0.025

q_L = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q_L = h_{fg = 2400 kJ/kg

now, at  h_{fg  = 2400 kJ/kg from saturated water tables;

T_L = 40 + ( 45 - 40 ) ( \frac{2400-2406.0}{2394.0-2406.0}\\} )

T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

T_L = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

4 0
3 years ago
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