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3 years ago

Which label on the wheelbarrow shows the pivot?

Physics

2 answers:

sdas [7]3 years ago

3 0

I think the answer is D.

Minchanka [31]3 years ago

3 0

The answer is D :)))))

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Two equally charged spheres of mass 1.00 g are placed 2.00 cm apart. When released, the initial acceleration of each sphere is 2

shusha [124]

Answer:

$q = 0.107 \mu C$

Explanation:

As we know that net force is given by

$F = ma$

here we have

m = 1.00 g = 0.001 kg

also we know that acceleration is given as

$a = 256 m/s^2$

now force is given as

$F = 0.001(256) = 0.256 N$

now by the formula of force we know that

$F = \frac{kq_1q_2}{r^2}$

$0.256 = \frac{(9\times 10^9)q^2}{(0.02)^2}$

now for solving charge we have

$q = 0.107 \mu C$

5 0

3 years ago

if the force of friction is acting on the sliding crate is 100 N how much force will be applied to maintain a constant velocity?

djverab [1.8K]

"Constant velocity" means zero acceleration, which means zero net force. So there must be100N pulling on the crate to cancel the 100N of friction force.

5 0

4 years ago

A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p

miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

The potential on the second plate

$V_{2}=V-V_{1}$

$V_{2}=51.5 -9.18$

$V_{2}=42.32\ v$

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0

3 years ago

A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is supported by an upward force o

Rainbow [258]

Answer:

5..No work is being done.

Explanation:

Hello!

Remember that the concept of work is defined as the force required to move a certain body distance, it is calculated as the product of force by distance.

therefore, the work required to raise the box is 150J.

However, the work required to keep the box lifted without moving is zero since although the box has a force due to the weight it does not move.

6 0

3 years ago

Is a solid PETROLEM

bearhunter [10]

No petroleum is a liquid

3 0

3 years ago

Read 2 more answers