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3 years ago

In which position would an object have a greater amount of potential energy

Physics

2 answers:

valentinak56 [21]3 years ago

7 0

Answer:

At the maximum height

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

Mathematically, it is given by:

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object relative to the ground

From the formula, we see that the potential energy is directly proportional to the height of the object above the ground: therefore, the larger the value of h, the larger the potential energy of the object.

Therefore, the object has the maximum potential energy when it is at the highest position above the ground.

Ludmilka [50]3 years ago

6 0

Answer: hanging 20 m from the ground

Explanation:

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Please hurry! 30 points

Cloud [144]

Answer:

50.000

Explanation:

4 0

3 years ago

A 190 g air-track glider is attached to a spring. The glider is pushed in 8.6 cm against the spring, then released. A student wi

spin [16.1K]

Answer:

The spring constant = 9.25 N/m

Explanation:

The equation of an object attached to a spring that is oscillating is

T = 2π√(m/k)

Where T = period of the oscillation, m = mass of the object, k = spring constant.

Making k the subject of the equation,

k = 4π²m/T²......................... Equation 1

Note: Period(T) is the time taken to complete one oscillation

Given: T = t/10 = 9.0/10 = 0.9 s, m = 190 g = 0.19 kg.

Constant: π = 3.14

Substitute these values into equation 1.

k = 4(3.14)²(0.19)/0.9²

k = 7.4933/0.81

k = 9.25 N/m

Thus the spring constant = 9.25 N/m

5 0

3 years ago

When two atoms share electrons the bond is ______

Ksju [112]

I think it is equal for your question

3 0

3 years ago

Read 2 more answers

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

lara [203]

Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

4 0

3 years ago

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}\times 9.8 \times 10^2$

$y = \frac{1}{2}\times 9.8 \times 100$

$y = 490 m$

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

$speed = \frac{distance}{time}$

$v = \frac{650}{10}$

$v = 65 m/s$

4 0

3 years ago