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Aloiza [94]
1 year ago
10

An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable i

n this experiment is a a. discrete random variable. b. categorical random variable. c. complex random variable. d. continuous random variable.
Physics
2 answers:
faust18 [17]1 year ago
8 0

The random variable in this experiment is a Continuous random variable.

Option D

<u>Explanation</u>:

The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.

Alja [10]1 year ago
7 0

Answer:

The random variable in this experiment is a Continuous random variable.

Explanation:

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If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should
liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

v_x = 14 cos30 = 12.12 m/s

v_y = 14 sin30 = 7 m/s

vertical distance between them

\delta y = 4 m

now we will use kinematics in order to find the time taken by the ball to reach at Allison

\delat y = v_y *t + \frac{1}{2} at^2

here acceleration is due to gravity

a = 9.8 m/s^2

now we will have

4 = 7 * t + \frac{1}{2}*9.8 * t^2

now solving above quadratic equation we have

t = 0.44 s

now in order to find the horizontal distance where ball will fall is given as

d = v_x * t

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

d = 12.12 * 0.44 = 5.33 m

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0
2 years ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
1 year ago
Current X is 2.5 A and runs for 39 seconds. Current Y is 3.8 A and runs for 24 seconds. Which current delivered more charge, and
Aleonysh [2.5K]

Answer: B. Current x delivered 6.3 C more then Y

Explanation:

7 0
2 years ago
A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
1 year ago
An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas
Brrunno [24]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0
2 years ago
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