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tangare [24]
3 years ago
14

A support crossbeam on a high speed train is made from a titanium rod that has a length of 0.650m when measured by an observer o

n the train. The beam is positioned such that the observer on the train measures an angle of 39degrees between the crossbeam and the floor of the train. (Figure 1) At what speed u would the train have to be traveling in order for an observer standing outside the train to measure the angle as 83degrees ? In the figure, S indicates the reference frame of the observer standing outside the train and S' indicates the reference frame of the observer on the train.
The beam has two dimensions: a height that is perpendicular to the floor of the train and a width that is parallel to the floor of the train. The angle between the beam and the floor appears different as measured by observers on the moving train and on the platform for which of the following reasons?

A. The crossbeam is moving relative to the observer on the platform so the height appears contracted.
B. The crossbeam is moving relative to the observer on the platform so the width appears contracted.
C. The crossbeam is moving relative to the observer on the train so the width appears contracted.
D. The crossbeam is moving relative to the observer on the train so the height appears contracted.
Physics
1 answer:
Vilka [71]3 years ago
3 0

Answer:

A) The crossbeam is moving relative to the observer on the platform so the height appears contracted.

Explanation:

The observer on the train and the beam are in the same reference frame. That means observer on the train will measure the proper length of the beam not the contracted length . the observer is outside and the plank is in the moving system,it will appear to be moving.

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What name is given to the process that occurs when light passes from air to water?
-Dominant- [34]

Answer:

Refraction

Explanation:

5 0
3 years ago
Which of the following is NOT an example of a correct interaction between the
makkiz [27]

Answer:

WE NEED ANSWER CHOICES

Explanation:

3 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
4 years ago
To convert centimeters to kilometers, which conversion factors would you need?
Lelechka [254]

Answer:

The last option is the only correct one if you like to multiply

The second last option is good if you like to divide.

Explanation:

Each fraction in the last two options has a value of 1

example

dividing by 1

15 cm /(100 cm/ 1 m) = 0.15 m          0.15 m / (1000 m/ 1km) = 0.00015 km

and

multiplying by 1

15 cm(1 m / 100cm) = 0.15 m         0.15m(1 km/1000m) = 0.00015 km

only one of the two fractions in each of the top two options has a value of 1.

3 0
3 years ago
A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast
Andrej [43]
To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
6 0
3 years ago
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