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4 years ago

A train travels 97 kilometers in 4 hours, and then 63 kilometers in 1 hours. What is its average speed?

Physics

1 answer:

Alina [70]4 years ago

4 0

Given data

Train covers a distance

(S₁) = 97 Km, in time (t₁) = 4 hr

and then covers a distance

(S₂) = 63 Km, in time (t₂) = 1 hr

What is the average speed (v) = ?

The average speed of the train is the ratio of the total distance covered to the total time taken,

since total distance covered = 97 km + 63 km

= 160 Km

and total time taken = 4 hr + 1 hr

= 5 hours

Therefore,

average speed = total distance covered /total time taken

 = 160 km/ 5 hr 

 = 32 km/hr

The average speed of the train is 32 km/hr

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Cadmium zinc telluride, zinc cadmium sulfide

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3 years ago

The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o

WINSTONCH [101]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which $\mu k$ = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant $\theta = 30^{o}$ .

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

$\sum F_{y} = 0$

N - W = 0

N = W

or, N = 75 lb

Again, equilibrium condition in the vertical direction is as follows.

$\sum F_{x} = 0$

$T_{2} - F_{k}$ = 0

$T_{2} = \mu_{k} N$

= $0.4 \times 75 lb$

= 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

$\sum F_{x} = 0$

$T Cos (30^{o}) + T Cos (30^{o}) = T_{2}$

$2T Cos (30^{o}) = T_{2}$

or, T = $\frac{T_{2}}{2 Cos (30^{o})}$

= $\frac{30}{2 Cos (30^{o})}$

= $\frac{30}{1.732}$

= 17.32 lb

Now, we will calculate the output power of the motor as follows.

P = Tv

= $17.32 lb \times 6$

= $103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}$

= 0.189 hp

or, = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

3 0

3 years ago

Read 2 more answers

A 0.80 kg basketball traveling upward at 5.0 m/s impacts an 8.0 10 kg tennis ball traveling downward at 5.0 m/s. The basketball’

vlabodo [156]

To solve this problem we will apply the concepts related to the conservation of momentum. This can be defined as the product between the mass and the velocity of each object, and by conservation it will be understood that the amount of the initial momentum is equal to the amount of the final momentum. By the law of conservation of momentum,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_1$ = Mass of Basketball

$m_2$ = Mass of Tennis ball

$u_1$ = Initial velocity of Basketball

$u_2$ = Initial Velocity of Tennis ball

$v_1$ = Final velocity of Basketball

$v_2$ = Final velocity of the tennis ball

Replacing,

$(0.8)(0.5)+(0.1)(-5.0)=(0.8)(0.3)+(0.1)v_2$

Solving for the final velocity of the tennis ball

$v_2 = 11m/s^2$

Therefore the velocity of the tennis ball after collision is 11 m/s

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3 years ago

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A hot-air balloon is rising straight up with a speed of 1.30 m/s. A ballast bag is released from rest relative to the balloon wh

DiKsa [7]

Answer:

The time before the ballast hits the ground is 1.186918s

Explanation:

We assume that the upward direction is positive.

The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.

The balloon has a velocity (V0y) = 3m/s

When the bag is released, it has a height of 5.36m

To touch the ground it makes a displacement of y = - 5.36m

The motion of the bag can be described as followed:

y = V0t + 1/2 * at²

t² + (2V0/a)*t - (2y/a) = 0

We can solve this equation for t:

t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2

t = -(V0/a) ± √((V0/a)² + (2y/a))

In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)

t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))

t = 0.132653 ± 1.054266

t= 1.186918

The time before the ballast hits the ground is 1.186918s

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3 years ago

Concave lenses have

Scilla [17]

Answer:

A. negative; virtual.

Explanation:

A concave lens always forms a virtual erect and diminished image. The reason why is that the image is actually formed by the intersection of virtually extended refracted rays.

Hope this helps!

Please mark as brainliest if correct!

3 0

2 years ago