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Naya [18.7K]
2 years ago
11

Rita throws a ball straight up into the air and catches it at the same position from which she threw it. The ball has 18 J of me

chanical energy when it leaves her hand.
If no energy is lost due to friction, which statements about the energy of the ball are true? Check all that apply.

The potential energy at the top of the ball’s motion is 18 J.
The kinetic energy is less when the ball is thrown than when it is caught.
The kinetic energy increases as the potential energy decreases.
The kinetic energy decreases as the potential energy increases.
The total mechanical energy of the ball stays constant.
The mechanical energy decreases as the ball moves up and increases as the ball comes down.
Physics
2 answers:
Zina [86]2 years ago
7 0
The 1st,3rd,4th, and 5th ones are the correct answers
Mashutka [201]2 years ago
5 0
The correct statements are statements 1,3,4, & 5
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You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg
Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

3 0
2 years ago
A roller coaster travels 200 feet horizontally and then rises 135 feet at an angle of 30 degrees above the ground. What is the m
Delvig [45]
Say the initial point is (0,0)

The final point is

x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft

Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
8 0
2 years ago
An object on a number line moved from x = 15 cm to x = 165 cm and then
olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

v_{avg}=\frac{x_{all}}{t}

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0
2 years ago
Read 2 more answers
In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0
2 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
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