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3 years ago

A force that pushes or pulls is known as A. A reaction force. B. An expected force. C. A positive force. D. An applied force.

Physics

2 answers:

otez555 [7]3 years ago

7 0

Answer:

the answer is an applied force

Explanation:

Julli [10]3 years ago

6 0

I believe it’s D, an applied force.

A reaction force seems to refer to Newton’s third law, but is relatively vague to be the answer to this question.

An expected force isn’t a concept nor the name of any subject of forces that is taught within the physics textbooks.

A positive force refers to direction, a negative force can have less, equal, or even more magnitude than the positive force, thus it contradicts itself. As that’s still a force that can push or pull.

So I believe it to be D.

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3 years ago

Read 2 more answers

The total charge a battery can supply is rated in mA⋅hmA⋅h, the product of the current (in mA) and the time (in h) that the batt

Nat2105 [25]

Answer:

118800 seconds

Explanation:

Given :

Voltage, V = 1.2 V

Resistance, R = 22 Ω

Applying Ohm's law, we get

Voltage, V = IR

Current $I=\frac{V}{R}$

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3 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by