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Tasya [4]
1 year ago
15

a battery of 5v and internal resistance of 5 ohms is connected to a resistor of 20 ohms. calculate the value of terminal voltage

Physics
1 answer:
Alborosie1 year ago
7 0

Answer:

A bird flies 75m in 15s.Calculate its speed

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Question points)
Marysya12 [62]

At number 5 (choice E).

6 0
3 years ago
A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc
OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0
3 years ago
Read 2 more answers
Pls help Which one of the following is NOT acceleration
ch4aika [34]

Answer:

C

Explanation:

Am increase in velocity?

3 0
2 years ago
Read 2 more answers
When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
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