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3 years ago

an apple in a tree has a gravitational potential energy of 175J and a mass of 0.36g . how high from the ground is the apple

1 answer:

5 0

The equation for potential energy is denoted as;

Pe = mgh,

where m = the mass, g = acceleration due to gravity, and h = vertical height of the apple. We are given the units for everything but height, which is also what we are solving for. We can then algebraically rearrange our initial equation to solve for h;

h = (Pe)/(mg)

Plug in your given units, and solve!

Post-check:

h = Pe/mg

h = 175J/(0.36g)(-9.81m/s^2)

h = appr. 49.5 meters

Note: Potential energy is a vector quantity; the displacement of the apple will be a negative number, but the distance itself, a scalar quantity, will be the absolute value of that.

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What change will always result in an increase in gravitational force between two objects

tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

$F = G\frac{m_1m_2}{d^2}$

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0

3 years ago

If the horizontal component of a vector is 6 m/s and the vertical component is also 6 m/s, what is the resultant value of the ve

BlackZzzverrR [31]

Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.

6 0

3 years ago

In a concave mirror parallel rays falling on it convergs at

ella [17]

Answer:

1) In a concave mirror parallel rays falling on it converges at F and 2F.

Explanation:

Spherical mirrors can be used for magnification of images. There are basically two types of spherical mirrors and they are converging mirror and diverging mirrors. The converging mirrors are also termed as concave mirrors and its basic work is to converge or combine light rays coming from a larger distance to a single point. Mostly the light beams falling parallel to the principle axis of the concave mirror will be acting as parallel rays. And when these parallel rays fall on the mirror, the converging point can be the focal point of the mirror.

Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.

5 0

3 years ago

If NEPA charges 5k per kWh, what is the cost of

tiny-mole [99]

Answer:

24k

Explanation:

We multiply by 200V by 24

8 0

2 years ago

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

3 years ago