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3 years ago

Plzzzzzz help :(((((((((((

Physics

2 answers:

OlgaM077 [116]3 years ago

7 0

Answer:

4. a) F

b) E

c) B

d) D

Hope it helps

RoseWind [281]3 years ago

6 0

F E B D, I think this is right...

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On the graph of voltage versus current, which line represents a 2.0 Ω resistor?

Vikki [24]

Answer:

Explanation:

According to ohm's law V = IR where;

V i sthe supply voltage (in volts)

I = supply current (in amperes)

R = resistance (in ohms)

In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

R = 10.0/5.0

R = 2.0ohms

Since the slope of line B is equal to 2 ohms, this shows that the line B is the one that represents the 2ohms resistor.

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4 years ago

While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it

diamong [38]

Answer: 6s

Explanation:

Vs=32m/s speed at beginning of slowing down

Vf=0m/s stop speed

a= -6 m/s² acceleration

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Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

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3 years ago

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$