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Assoli18 [71]
3 years ago
5

A large truck breaks down out on the road and receives a push back to town by a small compact car.

Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

A True. It agrees with Newton's third law

3 True. The car pushes the truck and goes at constant speed

Explanation:

To examine the final answers, we must silver the solution of the problem, if we see Newton's third law, action and reaction, we see that the car pushes the truck the action, the truck must oppose this force with a force applied on the car of equal magnitude, but opposite direction

In view of the above, let's review the statements.

A True. It agrees with Newton's third law

B False. Violate Newton's third law

C False  violate Newton´s third law

D False. The force is exerted by the car not specifically by the engine

E Faults if no force is exerted the truck should stop due to friction

Second question

If the two vehicles move at the same speed, the resulting force on each of them must be zero

1 False. If the truck doesn't get nine, it can't go at cruising speed

2 False if the car is accelerating it cannot go at constant speed

3 True. The car pushes the truck and goes at constant speed

4 False. If the truck slows it should slow down

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The disturbance does not have a specific motions 
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3 years ago
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If a battery is running down, a light bulb in the circuit will become dimmer because when voltage drops
lana [24]
A. There is less current flowing through the bulb.

Ohms law dictates that Electrical voltage = current x resistance. To understand current you would divide voltage by resistance. The less voltage the less current will run through the circuit.

Mark as brainliest if satisfied.
6 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
A piece of metal will feel colder than a piece of wood at the same temperature. Why?
alexira [117]
Metals in general, are good heat conductors 
7 0
3 years ago
HELPPP PLEASEEE!!!!!
devlian [24]

Answer:

1) The speed of sound increases

2)  440 Hz

3)  29°C

4)  17°C

5) 434 Hz

6)  12 m/s

7)  17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0
3 years ago
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