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2 years ago

How much time would it take a train locomotive running at 6000 Horsepower to use the

Physics

1 answer:

asambeis [7]2 years ago

7 0

Answer:

at least a day

Explanation:

Because when you have 6000 horsepower and 132000000 joules of energy it will take a little bit of time.

It would also be easier if you had the miles per hour

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What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

5 0

2 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.

motikmotik

Answer:

5 i think

Explanation:

4 0

3 years ago

A car is traveling with a velocity of 5.5 m/s and has a mass of 1200 kg. What is the kinetic energy?!

vesna_86 [32]

Answer:

Explanation:

The kinetic energy of the car can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the Mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 1200 \times {5.5}^{2} \\ = 600 \times 30.25$

We have the final answer as

Hope this helps you

3 0

2 years ago

A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran

marishachu [46]

<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train = 4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0

3 years ago