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3 years ago

HELP PLEASE!! 15 points awarded!

Physics

1 answer:

vekshin13 years ago

5 0

Answer:

its b

Explanation:

cause im in 7th grade ive been asked the same question

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An object slides down a frictionless inclined plane. At the bottom, it has a speed of 9.80 m/s. What is the vertical height of t

Papessa [141]

The vertical height of the given plane is 4.9 m.

The given parameters:

<em>speed of the object at the bottom of the ramp, v = 9.8 m/s</em>

The vertical height of the plane is calculated by applying principle of conservation of mechanical energy as follows;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{9.8 \times 9.8}{2 \times 9.8} \\\\h = 4.9 \ m$

Thus, the vertical height of the given plane is 4.9 m.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

3 years ago

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:

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

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

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

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

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.5 n blo

Naya [18.7K]

<Continuation of the question>

(a) the spring constant (in N/m) and

(b) the weight of the block hanging from the third spring.

the distance for the first spring is 20cm, the second 35cm, the third 50cm.

Answer:

a) To find the spring constant, we'll use the formula

F=kx, if we make k the subject we'll get

k=F/x, where F=4.5N, x = 35cm - 20cm = 15cm = 0.15m

k=F/x = 4.5N/0.15m = 30N/m is the Answer

b) to find the weight of block hanging on third spring

we use the formula F=kx

where k = 30N/m, x=50cm-20cm=30cm=0.30m

F=kx = (30N/m)*(0.30m) = 9N is the Answer

4 0

3 years ago

A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g o

kodGreya [7K]

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

$(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0$

$150*0.385 *( T - (-50)) + 200*0.450*(T - 120) + 300*2.46 * (T -20) = 0$

$57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0$

$57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0$

$57.75T + 0.90T+738T = - 2887.5 + 108+14760$

$796.65T= 11980.5$

T = 15.03°C

4 0

3 years ago

The spring constant in a given oscillating mass spring may be changed by

tia_tia [17]

(D) None of the above.

5 0

3 years ago