A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

Technician A says the counter gear is actually several gears machined out of a single piece of steel. Technician B says the coun

dexar [7]

Answer:

Option c. (Both Technician A and B are correct)

Explanation:

A transmission system consists of 3 shafts. The input shaft, the counter shaft, and the main shaft. The clutch gear always rotates with input shaft and is a crucial element of the input shaft.

The counter shaft is actually several gears machined out of a single piece of steel. The counter shaft may also be called counter gear or cluster gear. It is a secondary shaft that runs parallel to the mainshaft in a gearbox and is used to provide powers to machine components such as the drive axle.

The main gears (also called the speed gears) on main shaft (also known as the output shaft) are used to transfer rotation from counter shaft to the output shaft.

Hence in the light of above description, both technician A and B are correct.

4 0

3 years ago

Read 2 more answers

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

3 years ago

Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward

Dafna11 [192]

Answer:

0.54

Explanation:

Draw a free body diagram. There are 5 forces on the desk:

Weight force mg pulling down

Applied force 24 N pushing down

Normal force Fn pushing up

Applied force 130 N pushing right

Friction force Fnμ pushing left

Sum of the forces in the y direction:

∑F = ma

Fn − mg − 24 = 0

Fn = mg + 24

Fn = (22)(9.8) + 24

Fn = 240

Sum of the forces in the x direction:

∑F = ma

130 − Fnμ = 0

Fnμ = 130

μ = 130 / Fn

μ = 130 / 240

μ = 0.54

6 0

3 years ago

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0

3 years ago

A jeweler is determining the optical properties of an unknown blue gemstone. She uses an angle of incidence of 62°, and measures

Juliette [100K]

Answer:

n = 1.76

Explanation:

According to the rule of ( n1 sin theta1 = n2 sin theta2 )

we know both angles so we insert them to the law and apply n1 = 1

so 1/2 = n2 sin 62 and we get the final answer

3 0

3 years ago