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3 years ago

HELP ASAP! GIVING BRAINLIEST!

Physics

2 answers:

Blababa [14]3 years ago

8 0

Answer:

1. Emma standing on top of mountain

Since she is at the rest position and at some height from the ground so here energy is due to gravitational potential energy

So we have

gravitational potential energy

$U = mgH$

2. Emma jumping down from mountain top

Due to free fall Emma will start moving with some speed in downwards direction so here we have

$KE = \frac{1}{2}mv^2$

motion energy

3. tension in rope at Emma’s lowest position

Due to stretch in the rope here position come to the lowest end and speed comes to zero so whole energy is converted into elastic potential energy

$U = \frac{1}{2}kx^2$

elastic potential energy

4. Emma bouncing back

Due to bouncing back she will again have its kinetic energy with some speed upwards

$KE = \frac{1}{2}mv^2$

motion energy

expeople1 [14]3 years ago

6 0

Answer:

1. Emma standing on top of mountain

Since she is at the rest position and at some height from the ground so here energy is due to gravitational potential energy

So we have

gravitational potential energy

2. Emma jumping down from mountain top

Due to free fall Emma will start moving with some speed in downwards direction so here we have

motion energy

3. tension in rope at Emma’s lowest position

Due to stretch in the rope here position come to the lowest end and speed comes to zero so whole energy is converted into elastic potential energy

elastic potential energy

4. Emma bouncing back

Due to bouncing back she will again have its kinetic energy with some speed upwards

motion energy

Explanation:

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Can someone please help me with this question?

dexar [7]

The answer is B because, the farther the goes the speed starts to decrease and stops it depends on the frequency of the wave.

5 0

4 years ago

g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses

statuscvo [17]

Answer:

v₁f = 0.5714 m/s (→)

v₂f = 2.5714 m/s (→)

e = 1

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)

v₁f = 0.5714 m/s (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s (→)

e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1

It was a perfectly elastic collision.

8 0

3 years ago

The surface charge density on an infinite charged plane is −1.70 x 10^−6 C/m^2 . A proton is shot straight away from the plane a

goldfiish [28.3K]

Answer:

Acceleration of proton will be $-9.197\times 10^{12}m/sec^2$

Explanation:

We have given surface charge density of an infinite charged plate $\sigma = -1.70\times 10^{-6}C/m^2$

Electric field due to infinite sheet charge is given by $E=\frac{\sigma }{2\epsilon _0}=\frac{-1.7\times 10^{-6}}{2\times 8.85\times10^{-12}}=-0.096\times 10^6=-9.6\times 10^4N/C$

Charge in proton is given by $e=1.6\times 10^{-19}C$

So force on proton $F=qE=1.6\times 10^{-19}\times -9.6\times 10^4=-15.36\times 10^{-15}N$

Mass of proton $m=1.67\times 10^{-27}kg$

According to newtons second law force F = mass × acceleration

So $-15.36\times10^{-15}=1.67\times 10^{-27}a$

$a=-9.197\times 10^{12}m/sec^2$

3 0

3 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

icang [17]

Answer:

Part a)

$P = 5.72 kg m/s$

Part b)

$\Delta P = 2.93 kg m/s$

Part c)

$F = 44.4 N$

Part d)

$\Delta P = 5.02 kg m/s$

Part e)

$\Delta t = 0.113 s$

Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

$v = 11.8 cos29 \hat i + 11.8 sin29 \hat j$

$v_i = 10.3 \hat i + 5.72 \hat j$

Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

$P = 0.256(11.8)$

$P = 5.72 kg m/s$

Part b)

Change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(5.72 + 5.72)$

$\Delta P = 2.93 kg m/s$

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

$F = 44.4 N$

Part d)

Magnitude of change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(7.8 + 11.8)$

$\Delta P = 5.02 kg m/s$

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

$\Delta K = 0$

8 0

3 years ago

Firdavs [7]

1). The forces inside the atom are always, totally, completely, electrostatic forces. Those are so awesomely stronger than the gravitational forces that the gravitational ones are totally ignored, and it doesn't change a thing.

Parts 2 and 3 of this question are here to show us how the forces compare.

Part-2). The electrostatic force between a proton and an electron.

The constant in the formula is 9x10^9, and the elementary charge is 1.602 x 10^-19 Coulomb ... same charge on both particles, but opposite signs.

I worked through it 3 times and got 0.000105 N every time. So the best choice is 'C', even though we disagree by a factor of ten times. You'll see in part-3 that it really doesn't make any difference.

Part-3). Gravitational force between a proton and an electron.

The constant in Newton's gravity formula is 6.67x10^-11 . You'll have to look up the masses of the proton and the electron.

I got 2.163 x 10^-55 N ... exactly choice-C. yay !

Now, after we've slaved over a hot calculator all night, the thing that really amazes us is not only that the electrostatic force is stronger than the gravitational force, but HOW MUCH stronger ... 10^51 TIMES stronger. That's a thousand trillion trillion trillion trillion times stronger !

That's why it has no effect on the measurements if we just forget all about the gravitational forces inside the atom.

4 0

3 years ago

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