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sleet_krkn [62]
2 years ago
6

A girl is running toward the front of a train at 10 m/s. If the train is going 75 m/s on the Southbound tracks, what is the spee

d of the train relative to the girl?
Physics
2 answers:
sineoko [7]2 years ago
6 0

Answer:

velocity of the train with respect to girl is 85 m/s

Explanation:

As we know that girl is moving towards the train in front of it with speed

v_1 = 10 m/s

also we know that the speed of the train towards the girl is

v_2 = 75 m/s

Now we know that relative velocity of the train with respect to girl is given as

\vec v_{tg} = \vec v_t - \vec v_g

here let say the velocity of train is positive while girl is running opposite to train so its velocity is negative

\vec v_{tg} = 75 m/s - (-10 m/s)

\vec v_{tg} = 85 m/s

dimulka [17.4K]2 years ago
5 0
We can take as positive direction the direction of the train, and as negative direction the direction of the girl. The speed of the train relative to the girl will be given by the difference between the two velocities:
v' = v_t - v_g
which means
v'=75 m/s - (-10 m/s)=85 m/s
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4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco
MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

5 0
2 years ago
Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by
mrs_skeptik [129]
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R  as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
2 years ago
A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0
2 years ago
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
2 years ago
In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0
2 years ago
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