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3 years ago

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A red ball is thrown down with an initial speed of 1.1m/s from a height of 28 meters above the ground. Then 0.5 seconds after th

e red ball is thrown, a blue bar) is thrown upward with an initial speed of 24.6m/s, from a height of 1 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81m/s^2
1. What is the speed of the red ball right before it hits the ground?
2. How long does it take the red ball to reach the ground?
3. What is the maximum height the blue ball reaches?
4. What is the height of the blue ball 2 seconds after the red bal is thrown
5. How long after the red ball is thrown are the two balls in the air at the same height?

1 answer:

Nastasia [14]3 years ago

4 0

1.
v2 = u2 + 2gh

v2 = 1.12 + 2 * 9.8 * 28 = 550.01

v = 23.45 m/s

The speed of the red ball right before it hits the ground = 23.45 m/s.

2.

v = u + gt

23.45 = 1.1 + 9.8 * t

t = 2.28 seconds

The time it take the red ball to reach the ground = 2.28 seconds,

3.

Height above the ground = 0.8 m

v2 = u2 - 2gh

0 = 25.32 - 2* 9.8* h

h = 32.66 m.

The maximum height the blue ball reaches = (h + 1) = (32.66 + 1) = 33.66 m.

4.

Time of travel of the blue ball = (1.9 - 0.5) = 1.4 seconds.

s = ut - 0.5 gt2

s = 25.3* 1.4 - 0.5 * 9.8 * 1.42= 25.816 m.

The height of the blue ball 1.9 seconds after the red ball is thrown = (s + 1) = (25.816 + 1) = 26.816 m.

5.

Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.5) seconds.

Both the balls are at the same height :

28 - (s) = 1 + (h) ................{"s" & "h" are the displacements of the red & the blue ball respectively.}

28 - (ut + 0.5 gt2) = 1 + (ut - 0.5gt2)

28 - (1.1 t + 0.5 * 9.8 t2) = 1+ (25.3 (t-0.5) - 0.5*9.8*(t-0.5)2)

Now we have to solve the above equation to find the time after which both the balls are at the same height.

28 - 1.1t - 4.9t2 = 1 + 25.3t - 12.65 - 4.9t2 + 4.9 t - 1.225

(28 - 0.8 + 12.65 +1.225) = (25.3 + 4.9 + 1.1) * t

t = 41.075 / 31.3 = 1.3123 = 1.31 seconds (approx.)

The time after the red ball is thrown are the two balls in the air at the same height = 1.31 seconds.

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maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

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x = (2.9 ± 0.3) m

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The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

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The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

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natka813 [3]

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With the given values of $F,G,m_1,r$ , we have

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Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

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$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

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$m_2=7.3\times10^{22}\,\mathrm{kg}$

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1 year ago

Read 2 more answers

A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte

zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

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