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4 years ago

A ball is thro.d hit the grouvelocity)?

Physics

1 answer:

kati45 [8]4 years ago

3 0

Answer:

The velocity of the ball is 3.52 m/s.

Explanation:

A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.

Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;

u = $\sqrt{\frac{2H}{g} }$

Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/ $s^{2}$ , then;

u = $\sqrt{\frac{2*61.9}{10} }$

= $\sqrt{\frac{123.8}{10} }$

u = 3.5185

The velocity of the ball is 3.52 m/s.

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In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro

blondinia [14]

In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.

Ok, Rameses:

Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton

Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg

a). The force between two charges is

F = (9 x 10⁹) Q₁ Q₂ / R²

= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

      = ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

=    8.05 x 10⁻⁸ Newton .

b). Centripetal acceleration =

   v² / r .

A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

   =    7.7 x 10²² m/s² .

That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .

I'm going to leave that step for you.   Good luck !

4 0

3 years ago

Which of the following is true about inertia

Tamiku [17]

Answer:

The correct answers are It is the resistance of an object to changes in its motion, and It is a force

3 0

3 years ago

You have designed a machine that requires 1000 J of work from a motor for every 800 J of useful work done by the machine. What i

valentinak56 [21]

Answer:

80%

Explanation:

$efficiency = \frac{useful \: work \: done}{total \: energy \: input}$

800 / 1000 = 0.8

Efficiency = 0.8 *100 = 80%

4 0

2 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

ΔP.E = $-P.E_i$

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

$\Delta U =(0.70) (490.5)$

ΔU = 343.35 J

Thus, the magnitude of the increase is = 343.35 J

7 0

3 years ago