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xenn [34]
3 years ago
13

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

Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

<h2>line B</h2>

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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The upward force on an airplane's wing is thrust.
kompoz [17]

No.  'Thrust' is what most people in aviation call the force
that pushes the aircraft forward. 

The same people generally call the upward force on the wing "lift".

3 0
3 years ago
A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a sp
fgiga [73]

Answer:

Explanation:

Change in length of spring = 2.13 m

Component of weight acting on spring = mg sinθ

so

mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.

Here x = 2.13

mg sin17 = k x 2.13

31 x 9.8 sin17 = k x 2.13

k = 41.7 N/m

b ) In case surface had friction , spring would have stretched by less distance .

It is so because , the work done by gravity in stretching down is stored as potential energy in  spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc  so spring stretches less.

5 0
3 years ago
. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0
2 years ago
Which graph showing constant acceleration correctly places the independent and dependent variables?
svp [43]
Do you have a picture
6 0
3 years ago
Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False
kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

\vec{L} = I \vec{\omega}

where I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2 is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2} so \vec{\omega} \neq 0 unless \vec{r} ║  \vec{v}.

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0
2 years ago
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