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alexgriva [62]
3 years ago
6

What is question 22?

Physics
2 answers:
zalisa [80]3 years ago
5 0
The object moves 18 meters in 6 seconds (10 - 4).

18m/6s= 3m/s

Answer "A"
Fynjy0 [20]3 years ago
4 0

the answer is a!! its pretty simple I just read the graph.
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A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass
arlik [135]

Answer:

a) v_{60^{o}} =4.98 m/s

b) \theta_{max}=79.34^{o}

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

K_{lowest}=K_{60^{o}}+U_{60^{0}}

So now, we can substitute the respective equations for kinetic and potential energy so we get:

\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}

we can divide both sides of the equation into the mass of the pendulum so we get:

\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}

and we can multiply both sides of the equation by 2 to get:

v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}

so we can solve this for v_{60^{o}}. So we get:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

cos \theta = \frac{y}{4m}

so we solve for y to get:

y = 4cos \theta

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

h_{60^{o}}=4m-2m

h_{60^{o}}=2m

So now we can substitute the data in the velocity equation we got before:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}

so

v_{60^{o}}=4.98 m/s

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

K_{lowest}+U_{lowest}=K_{max}+U_{max}

In this case, we know that U_{lowest} will be zero and K_{max} will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

K_{lowest} =U_{max}

And now we substitute for the respective kinetic energy and potential energy equations.

\frac{1}{2}mv_{lowest}^{2}=mgh_{max}

again, we can divide both sides of the equation into the mass, so we get:

\frac{1}{2}v_{lowest}^{2}=gh_{max}

and solve for the height:

h_{max}=\frac{v_{lowest}^{2}}{2g}

and substitute:

h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}

to get:

h_{max}=3.26m

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

cos \theta_{max}= \frac{0.74}{4}

\theta_{max}=cos^{-1}(\frac{0.74}{4})

so we get:

\theta_{max}=79.34^{o}

5 0
3 years ago
Kilo-  -  1,000<br>deka-  -  10<br><br>How much larger is a kilo- than a deka-?
Ilya [14]
lets's \ take \ an \ example \ one \ gram: \\ \\ 0,001kg \\ \\ 0,01hg \\ \\ 0,1dag \\ \\ \boxed{1g} \\ \\ 10dg \\ \\ 100cg \\ \\ 1000mg  \\ \\ the \ units \ are \ rising \ and \ decreasing \ from \ 10 \ to \ 10 \ results \ \\ \\ a \ kilogram \ is \ 100 \ times \ larger \ than \ a \ dekagram
6 0
3 years ago
Two resistors of 2 ohms each are connected in parallel, another resistor of 1 ohm is connected in series with the parallel combi
insens350 [35]

Answer:

The power dissipated in either one of the parallel resistors is 2 V

Explanation:

Given;

two parallel resistors, R₁ and R₂ = 2 ohms

The total resistance of the Two resistors of 2 ohms connected in parallel is;

R_T = \frac{R_1R_2}{R_1+R_2} = \frac{2*2}{2+2} = \frac{4}{4} = 1 \ ohm

when connected to another resistor  of 1 ohm in series, the total resistance becomes;

Rt = R₁ + R₂

Rt = 1 + 1 = 2 ohms

Current in the circuit, I = voltage / total resistance

                                      = 2 /2 = 1 A

the overall circuit has been resolved to series connection, and current flow in series circuit is constant.

Power = I²R

Thus,  power dissipated in either one of the parallel 2 ohms resistors is;

Power = I²R = (1)² x 2 = 2 V

3 0
4 years ago
Can someone help me with science:
Oliga [24]

Answer:

trur

Explanation:

gvidttcyhjddgdhjfjdjsjs

6 0
3 years ago
Read 2 more answers
What causes the balloon to move along the string?
ExtremeBDS [4]
Gravity? Im almost sure thats it
3 0
3 years ago
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