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3 years ago

A 2.0 kg mass is lifted 4.0 m above the ground. find the change in gravitational energy

Physics

1 answer:

Makovka662 [10]3 years ago

6 0

Answer:

78.4 J

Explanation:

Given:

Mass of the object (m) = 2.0 kg

Height or displacement of the mass (h) = 4.0 m

Now, change in gravitational potential energy (ΔU) = ?

Change in gravitational potential energy of an object is the energy stored by the object when it is raised to some height above the reference ground level.

Change in gravitational potential energy of an object of mass 'm' raised to a height 'h' above the ground is given as:

$\Delta U=mgh$

Where, 'g' is the acceleration due to gravity and has a value of 9.8 m/s².

Now, plug in all the given values and solve for ΔU. This gives,

$\Delta U=(2.0\ kg)(9.8\ m/s^2)(4.0\ m)\\\\\Delta U=78.4\ J$

Therefore, the change in gravitational potential energy is 78.4 J.

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A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

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Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

(Either option A or B without the minus symbol before $F_{bat}$ , likely typographical error)

Reason:

Force exerted on the bat by the ball = $F_{ball}$

Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

Where the force of the bat is high, the ball is accelerated to travel at high speed

Learn more Newton's Laws of motion here:

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3 years ago

Ian walks 2 km to his best friends house, then walks 0.5 km to the library. He then makes 2.5 km walk home. The entire walk took

allochka39001 [22]

1. 2+0.5+2.5= 3. 2km/hr average

2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration

3. 15m/s divided by 2.5 = 6m/s acceleration

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4 years ago

Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)

Crazy boy [7]

Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100% = 1.25%

This is a percent error, we do not need to round this.

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3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

The number of fringe is z = 3 fringes

The ratio is $I = 0.2545I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$