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2 years ago

If a laptop of mass 1.4 kg is lifted to a table of height 0.8 m, how much gravitational potential energy is added to the laptop?

2 answers:

4 0

Potential Energy = mgh ,

mass = 1.4 kg, g ≈ 10 m/s², h = 0.8m

Potential Energy = mgh = 1.4 * 10 * 0.8 = 11.2 N

Potential Energy added = 11.2 N

Aleks [24]2 years ago

3 0

To calculate Potential Energy you need to Multiply 1.4*.8*9.8.... 9.8 is the acceleration due to gravity an they expect you to know to add 9.8 when you calculate Potential Energy. So the correct answer is 10.976 but rounded up is 11.

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The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

W = ΔK

Where work is defined as the product of force by distance

W = fr x cos 180

The angle is because the friction force opposes the movement

Δk = $K_{f}$ –K₀

ΔK = 0 - ½ m v₀²

We substitute

- fr x = - ½ m v₀²

fr = ½ m v₀²/x

8 0

2 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

2 years ago

Backpackers often use canisters of white gas to fuel a cooking stove's burner. If one canister contains 1.45 L of white gas, and

lorasvet [3.4K]

Answer: One canister contains 1.03 Kg of fuel,

Explanation:

The density is defined as the relation between the mass and the volume $p=\frac{m}{v}$

First of all you need to have the same units for volume and density, so:

For the volume,

$1.45 L * \frac{1.10*10^{-3}m^{3} }{1 L} = 1.45 *10^{-3} m^{3}$

For the density,

$p = 0.710\frac{g}{cm^{3} } *\frac{1cm^{3} }{1*10^{-6}m^{3} }*\frac{1 Kg}{1*10^{3}g } = 710\frac{Kg}{m^{3} }$

From the density equation we have $m=p*v$ , so:

$710\frac{Kg}{m^{3} } *1.45*10^{-3} m^{3} = 1.03Kg$

4 0

2 years ago

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$