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mihalych1998 [28]

2 years ago

12

Playing catch with a friend in a moving train. When you toss the ball in the direction the train is moving, how does the speed o

f the ball appear to an observer standing at rest outside the train?
I understand basic relative motion, just the ball moving on another moving object confuses me!

2 answers:

likoan [24]2 years ago

5 0

Well it would simply appear faster because the speed of the ball and the speed of the train.

telo118 [61]2 years ago

5 0

Answer:

The speed of the ball appear faster when we observe it from ground frame at rest.

Explanation:

As we know that by relate motion concept

$v_{AB} = v_A - v_B$

here when ball is thrown in the direction of train motion with velocity "v" relative to the train then we will assume here that let say the train at that moment is running at speed $v_T$

now by above concept of relative velocity we will have

$v = v_{ball} - v_T$

now for the speed of ball

$v_{ball} = v + v_T$

so from this equation we can see that the actual speed of ball in ground frame will be more than the speed observed in train frame

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2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

<em>weight of the load, w = 557 N</em>
<em>force applied , F = 410 N</em>
<em>angle of force, = 15°</em>
<em>coefficient of kinetic friction = 0.46</em>
<em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

<em>m is the mass of the recycling bin</em>
<em /> $F_k$ <em> is the frictional force </em>

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

2 years ago

How do l calculate e<br> nergy

aivan3 [116]

The formula for energy of motion is KE = .5 x m x v^2

Ke= Kinetic Energy in Joules

m = Mass in Kilograms

v = Velocity in Meters per Second

4 0

2 years ago

The net force experienced by an object is increased. What affect does this have on the acceleration of the object?

Evgesh-ka [11]

Answer:

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

Explanation:

hope it helps pls give me brainless

6 0

3 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

Help please! Will give brainliest to first correct answer!

Olegator [25]

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.

8 0

2 years ago