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3 years ago

8

You can increase the gravitational potential between a car and the road by

1 answer:

yarga [219]3 years ago

7 0

Answer:

D-Driving the car faster down the road.

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If you could go a year into the future, or 3 seconds into the past, where would you go? What date? I would go to June 6th, 2018

vova2212 [387]

Idk I would go back to pre-k

8 0

3 years ago

Is 5g greater than 20.43 mg

worty [1.4K]

yes that's true 6g is larger

4 0

3 years ago

Read 2 more answers

An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which

PolarNik [594]

Answer:

R=m*g-∀fl*g*l3

Explanation:

<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>

From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced

density of iron =mass/ volume

rho=m/l3

mass=rhol3

weight fluid=rhofluid*g*Volume

weight of fluid=rhofluid*g*l3

F=∀fl*g*l3

Downward force is weight of iron

w=m*g

Reading on the spring scale

R=w-F

R=m*g-∀fl*g*l3

m=mass of iron

g=acceleration due to ravity

rhfld=density of fluid

l3=volume of fluid displaced

6 0

3 years ago

Four people are spaced at different distances from where a dog is barking. The barking will sound the loudest to the person who

Volgvan

Answer:

Closest to the dog.

Explanation:

Sounds are louder the closer you are to them.

6 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago