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Ronch [10]
3 years ago
12

How does air resistance affect an object’s speed?

Physics
1 answer:
tatuchka [14]3 years ago
4 0
When air resistance<span> acts, acceleration during a fall </span>will<span> be less than g because </span>air resistance affects<span> the motion of the falling </span>objects<span> by slowing it down. </span>Air resistance<span> depends on two important factors - the</span>speed<span> of the </span>object<span> and its surface area. Increasing the surface area of an </span>object<span> decreases its </span>speed<span>.</span>
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Which change would decrease the total current, I, flowing through this circuit?
iren [92.7K]

Answer:

A. Increasing the voltage of the battery

Explanation:

The relationship between voltage, V, current, I and resistance, R, is given as follows;

V = I × R

∴ I = V/R

From the above relationship, the current flowing in the circuit is directly proportional to the voltage of the battery, and inversely proportional to the resistance, 'R', of the circuit

Therefore, increasing the voltage, 'V', of the battery, increases the total current, 'I', flowing in the circuit.

7 0
3 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

I =0.011 A

5 0
3 years ago
The SAF operates the M113 Ultra APC.
HACTEHA [7]

The volume of the object must be no larger than 11.15 m^3.

Explanation:

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The density of water is:

\rho = 1000 kg/m^3

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

\rho = \frac{m}{V}

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m is the mass of the object

V is its volume

For the object in this problem, the mass is

m=1.115\cdot 10^4 kg

Therefore, we can re-arrange the equation to find its volume:

V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3

So, the volume of the object must be no larger than 11.15 m^3.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

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The answer would be true
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