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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.

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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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

Anon25 [30]

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>

This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

F is the applied force
v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

7 0

2 years ago

A spring is stretched 2.0 cm. If the same spring is stretched 8.0 cm the ratio of the second and first potential energies of the

jok3333 [9.3K]

Potential energy of spring equals K times X squared divided by 2 where X is displacement

4 times squared equals 16

choose 1st answer

6 0

3 years ago

Which is not a common extrusive igneous rock

KonstantinChe [14]

Granite is the answer i think

8 0

3 years ago

Driving safely at night requires seeing well not only in low light conditions, but also being able to see low contrast

Monica [59]

Answer:

True

Explanation:

Driving safely at night requires seeing well not only under low light, but also requires drivers to see low-contrast objects.

8 0

2 years ago