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Tom [10]
3 years ago
9

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

Physics
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0            -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 mVf^{2}= 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 mVi^{2}=0.5×200×Vi^{2}=100Vi^{2} j

Put these values in eq 1, we get;

0-100 Vi^{2}+3018.4-0=0

-100 Vi^{2}=-3018.4

==> Vi^{2}= \frac{3018.4}{100} = 30.184

==>  Vi = \sqrt{30.184}  = 5.5 m.sec

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2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground  it has some gravitational potential energy with respect to hits the ground  

Formula for gravitational potential energy given by  

Potential Energy = mgh

Where ,

m = mass  

g = acceleration due to gravity  

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

Potential Energy = 0.375\times9.8\times1.37

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

Potential Energy = 0.375\times0.67\times9.8

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0
3 years ago
An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma
Jet001 [13]

Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

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If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec
Phantasy [73]

A graph of real speed can have a section that's as steep as you want,
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Any vertical line on a graph, even it it's only a tiny tiny section, means
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It also means that the speed took no time to change from one value to
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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
ZanzabumX [31]

Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

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3 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

Answer:

Moment of inertia of the solid sphere:

I

s

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2

5

M

R

2

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(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

M

M

4 0
3 years ago
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