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3 years ago

How does the force of gravity affect the rate of acceleration?

1 answer:

3 0

Gravity lets all objects fall to the ground at the same speed, 9.8 m/s/s. If the force of gravity were stronger, such as 10 m/s/s, the rate of acceleration would be faster.

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought

IgorLugansk [536]

Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:

(9 * 10^-12) / 3 = 3 * 10^-12 C

So the charge in each ball will be 3 * 10^-12 C

8 0

3 years ago

We use electrical devices leopard deuce motion, light, and sound. Which aspect of energy explains why these devices are possible

poizon [28]

Answer:

B) Energy can change from one form to another

Explanation:

It goes from electrical energy from the mains into motion, light and sound

3 0

3 years ago

Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5

andrezito [222]

Answer:

R = 1.2295 10⁵ m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

how angles are measured in radians

θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

$\frac{y}{R} = 1.22 \frac{ \lambda}{D}$

R = $\frac{ y \ D}{1.22 \lambda}$

let's calculate

R = $\frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}$

R = 1.2295 10⁵ m

3 0

2 years ago

Definition of AM and Fm Wave? in your own word

vfiekz [6]

Answer:

In AM broadcasting, the amplitude of the carrier wave is modulated to encode the original sound. In FM broadcasting, the frequency of the carrier wave is modulated to encode the sound. A radio receiver extracts the original program sound from the modulated radio signal and reproduces the sound in a loudspeaker.

8 0

3 years ago