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3 years ago

How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?

Physics

2 answers:

Dovator [93]3 years ago

8 0

Answer:

Answer: B. 196 J

• potential energy is calculated from the formula:

${ \tt{p.e = mgh}}$

• m = 10 kg

• g = 9.81 m/s²

• h = 2m

→ Substitute:

${ \tt{p.e = (10) \times (2) \times (9.81)}} \\ \\ → \: { \boxed{ \tt{potential \: energy = 196.2 \: \: joules}}}$

Lorico [155]3 years ago

6 0

Answer:

D. 196 J

Explanation:

PE=mgh=10×9.8×2=196J

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You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

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3 years ago

Look at the graph. What is the slope of the line?

NeTakaya

For the first hour, the slope is zero.

After that, the slope is -2 miles per hour.

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3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

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3 years ago

You are traveling at 14 m/s for 20 seconds. What is your displacement?

enyata [817]

Velocity = 14 m/s

Time = 20 s

Displacement = Velocity×Time = (14×20) m = 280 m

The displacement is 280 m towards the direction of motion.

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3 years ago

Read 2 more answers

What us the difference between accurate data and reproducible data

Nikolay [14]

It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!

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4 years ago