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Rama09 [41]
3 years ago
10

3. Two Metra trains approach each other on separate but parallel tracks. Train A has a speed of 90 km/ hr, train B has a speed o

f 80 km/ hr. Initially, the two trains are 2.71 km apart. How long will it take the two trains to meet?
a. Write two complete equations that describe the position of: Train A:
Train B:
c. At what exact time will train A and train B be at the same position?
d. At what position is train A when it meets train B?
e. How far has train B travelled in this time?

CAN SOMEONE PLEASE HELP ME I HAVE BEEN TRYING TO SOLVE FOR 5 HOURS AND I STILL DONT KNOW THE ANSWER
Physics
1 answer:
Elden [556K]3 years ago
8 0

Answer:he trains take 57.4 s to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

The relative velocity of the train A with respect to B is given by,

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

Substitute 2.71 km for d and 170 km/h for

Express the time in seconds.

Thus, the trains cross each other in 57.4 s.

Explanation:

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What is the average acceleration during the time interval 0 seconds to 10 seconds?
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Answer:

yea its D  .

Explanation:

3 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
3 years ago
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
Anon25 [30]

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

4 0
3 years ago
Extremely large main sequence stars consume their fuel quickly and burn hot and bright. As they consume all their hydrogen, they
beks73 [17]

Answer:  Their temperature decreases dramatically, but their luminosity increases only slightly.

Explanation: Edmentum answer

3 0
3 years ago
4. A hiker leaves camp and using a compass walks 4 km East, 6 km South, 3 km East, 5 km North, 10 km West, 8 km North and 3 km S
PSYCHO15rus [73]

Answer:

Because the hiker walked directly west and then directly north the two legs of the hike forms a right triangle. Therefore we can use the Pythagorean Theorem to solve this problem.

 

c=5

The hiker is 5km from camp and should head in a generally south-east direction

7 0
3 years ago
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