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3 years ago

On a cold winter day, the flow of heat is from the outside in. A.True B.False

Physics

2 answers:

fredd [130]3 years ago

8 0

Your answer s obviously false

lbvjy [14]3 years ago

7 0

Answer:

False

Explanation:

ooga booga, cold winter day cold outside, heat not outside ooga booga.

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What is the value of work done on an object when a

tino4ka555 [31]

W = F * s
Here, F = 50 N
s = 15 m

Substitute their values,
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!

7 0

2 years ago

A flashlight contains a battery of two cells in series, with a bulb of resistance 12 Ohms. The internal resistance of each cell

Elina [12.6K]

Answer:

1.5024

Explanation:

Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.

R = 0.26 + 0.26 + 12 = 12.52

The bulb has a voltage of 2.88 volts across it. You can get the current from that.

i = E / R

i = 2.88 / 12 =

i = 0.24 amps.

Now you can get the voltage drop across the two cells.

E = ?

R = 0.26

i = 0.24 amps

E = 0.26 * 0.24

E = 0. 0624

Finally divide the 2.88 by 2 to get 1.44

Each cell has an emf of 1.44 + 0.0624 = 1.5024

4 0

2 years ago

A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi

KatRina [158]

Answer:

Explanation is given

Mark Me as Branliest

3 0

2 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago

Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. Th

Dima020 [189]

Answer:

Explanation:

Let the thickness of the film is t and the refractive index of the material of film is n.

When light travels through a sheet of thickness t, the optical path traveled is nt.

When the path of one of slit is covered by a sheet of thickness t, the optical path becomes

x = ( n - 1) t

As the one fringe is shift, so the optical path changed by one wavelength.

i.e., x = λ

So, λ = ( n - 1) t

$t=\frac{\lambda }{n-1}$

7 0

2 years ago