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2 years ago

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A system in which only one particle can move has the potential energy shown in (figure 1). Suppose u1 = 60 j. What is the y-comp

onent of the force on the particle at y = 0. 5 m ?.

1 answer:

marta [7]2 years ago

3 0

The y-component of the force on the particle at the given position is 120 N.

<h3>Electric force on the particle</h3>

The electric force on the particle is determined by applying Coulomb's law and work-energy theorem as shown below;

Fd = W

Where;

F is the applied force
d is the distance
W is potential

F = W/d

F = 60/0.5

F = 120 N

Thus, the y-component of the force on the particle at the given position is 120 N.

Learn more about electric force here: brainly.com/question/20880591

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A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l

lianna [129]

For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.

3 0

3 years ago

tatuchka [14]

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

$F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N$

2) $10^{\circ}$

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

$\theta = tan^{-1}(\frac{F_x}{F_y})$

where

$F_x$ is the force in the horizontal direction

$F_y$ is the force in the vertical direction

In this problem,

$F_x = F_w = 1.68 N$

$F_y = W = 9.42 N$

and so we find:

$\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}$

4 0

3 years ago

The half-life of a certain isotope is 15 minutes. How much of a 400 g sample will remain after 90 minutes?12.5 g6.25 g26.7 g66.7

Virty [35]

There is a total of 6 half lives that need to take place.

ONE HALF LIFE = 200

TWO HALF LIFES = 100

THREE HALF LIFES = 50

FOUR HALF LIFES = 25

FIVE HALF LIFES = 12.5

SIX HALF LIFES = 6.25

The answer is 6.25g

4 0

3 years ago

aivan3 [116]

Answer:

c is trueeeeeeeeeeeeeee

7 0

3 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

4 years ago