Ask question

Ask question

All categories

3 years ago

10

A .422 kg coffee mug rests on a table top 0.63 m above the floor. What is the potential energy of the mug with respect to the fl

oor?

1 answer:

kondaur [170]3 years ago

4 0

Answer:

2.61 J

Explanation:

Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.

Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.

We compute U

U = mgy

= 0.422 kg × 9.8 m/s² × 0.63 m

= 2.605 J

≅ 2.61 J

So, the potential energy of the mug with respect to the floor is 2.61 J

You might be interested in

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

Write a 150 word paragraph or two that describes at least three everyday things that exist or occur because of science. Make sur

Andreyy89

almost everything artafitial would have been made because of science. However nothing natural would exist because of science. All science does is look around and see what's up.

3 0

3 years ago

Read 2 more answers

A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

8 0

3 years ago

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

julia-pushkina [17]

Answer:

$5.5\cdot 10^{-11} C$

Explanation:

The capacitance of the parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=190 mm^2 = 190 \cdot 10^{-6} m^2$ is the area of the plates

$d=1.2 mm = 0.0012 m$ is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F$

The energy stored in the capacitor is given by

$U=\frac{Q^2}{2C}$

Since we know the energy

$U=1.1 nJ = 1.1 \cdot 10^{-9} J$

we can re-arrange the formula to find the charge, Q:

$Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C$

8 0

3 years ago

What is the momentum of a 750 kg car traveling at a velocity of 25 m/s north?

olga_2 [115]

Answer:

18750 kg-m/s

Explanation:

Momentum = mass x velocity

3 0

3 years ago

Read 2 more answers