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nata0808 [166]
2 years ago
12

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

e boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance?.
Physics
1 answer:
zlopas [31]2 years ago
4 0

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0  + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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Answer:

acceleration of the car is 3 m\s^2

Explanation:

from rest means the initial velocity (vi) is zero

time = 5s

final velocity (vf) = 15m\s

a = vf - vi \ t

a = (15-0) \ 5

a= 3 m\s^2

which means that the car is speeding up 3 meters every second

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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

MV=mv

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

V = \frac{mv}{M}

Also we have that m/M = 1/7 times

On the other hand we have from law of conservation of energy that

W_f = KE

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

F_f*S = \frac{1}{2}MV^2

\mu M*g*S = \frac{1}{2}MV^2

\mu g*S = \frac{1}{2}( \frac{mv}{M})^2

\mu = \frac{1}{2} (\frac{1}{7}v)^2

\mu = \frac{1}{98}v^2

\mu = \frac{1}{g(98)(5.1)}v^2

Here we can apply the law of conservation of energy for light mass, then

\mu mgs = \frac{1}{2} mv^2

Replacing the value of \mu

\frac{1}{g(98)(5.1)}v^2  mgs = \frac{1}{2}mv^2

Deleting constants,

s= \frac{(98*5.1)}{2}

s = 249.9m

7 0
3 years ago
Why do we use scatter plots for most data in Physical Science? Question 1 options: because the teacher said to use scatter plots
SVEN [57.7K]

Answer:

scatter plots show the relationship between the independent and dependent variables

Explanation:

A scatter plot is a graph which shows two variables plotted along two axes (usually the x and y axes). Scatter plots are useful in establishing any form of correlation between the dependent and independent variables in any study.

Correlation simply means the degree of relationship between variables, that is, how much does one variable affect the other? When scatter plots are almost a straight line graph, there is a high correlation between the variables. When the points in a scatter plot are isolated, there is little (sometimes zero) correlation between the variables.

8 0
3 years ago
The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a
rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0
3 years ago
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