The type of brightness in which all

Kenny wants to get to Washington DC within 4 hours. Washington DC is 133 miles away from where he is. What is the avg speed he m

umka21 [38]

He must travel 35 mph

4 0

2 years ago

#2

sladkih [1.3K]

Answer:

mass and distance

Explanation:

force is mass while motion can also be regard as distance or movement

3 0

3 years ago

Maggie participated in a race. She ran 100 meters west and then turned back and ran to the starting line to complete the race. M

earnstyle [38]

Her total distance was 200 meters, since she ran 100 meters west and the same 100 meters back. This gives an average speed of 200 m / 40 s = 5 m/s.

However, her displacement was 0, since she ended at the same place she began, therefore, her average velocity is 0 / 40 s = 0.

3 0

3 years ago

A conducting sphere with a radius of 0.25 m has a total charge of 5.90 mC. A particle with a charge of −1.70 mC is initially 0.3

notsponge [240]

Explanation:

The given data is as follows.

$r_{1}$ = 0.25 m, q = 5.90 mC = $5.90 \times 10^{-3} C$

$r_{2}$ = 0.35 m, q = 1.70 mC = $1.70 \times 10^{-3} C$

(a) Now, we will calculate the electric potential as follows.

V = $k \frac{q}{r}$

First, we will calculate the initial and final electric potential as follows.

$V_{i} = 9 \times 10^{9} \times \frac{5.90 \times 10^{-3}}{0.25 m}$

= $212.4 \times 10^{6}$

or, = $2.124 \times 10^{8}$

$V_{f} = 9 \times 10^{9} \times \frac{1.70 \times 10^{-3}}{0.35 m}$

= $43.71 \times 10^{6}$

or, = $4.371 \times 10^{8}$

Hence, the value of change in electric potential is as follows.

$\Delta V = V_{f} - V_{i}$

= $4.371 \times 10^{8} - 2.124 \times 10^{8}$

= $2.247 \times 10^{8}$ V

Therefore, the difference in electric potential energy is $2.247 \times 10^{8}$ V.

(b) Now, we will calculate the potential energy as follows.

P.E = qV

= $-1.70 \times 10^{-3}C \times 2.247 \times 10^{8}$ V

= $-3.8199 \times 10^{5}$

Therefore, the change in the system's electric potential energy is $-3.8199 \times 10^{5}$ .

4 0

3 years ago

A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: 

Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is:

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
W = 1250π ∫ x dx
W = 625π * (x2 – x1)

x2 = 14 ft + 15 ft = 29 ft

x1 = 14 ft + 1 ft = 15 ft

W = 625π * (29^2 - 15^2)
W = 385,000π ft-lbs = 1,209,513.17 ft-lbs

3 0

3 years ago