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Elodia [21]
3 years ago
9

A 13.8 kg block and a 25.5 kg block are resting on a horizontal frictionless surface. between the two is squeezed a spring (spri

ng constant = 1131 n/m). the spring is compressed by 0.164 m from its unstrained length and is not attached permanently to either block. with what speed does each block move away when the mechanism keeping the spring squeezed is released and the spring falls away?
Physics
1 answer:
raketka [301]3 years ago
4 0
Some dogs may inherit a susceptibility to epilepsy.

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What happens to a circuit's resistance as more resistors are added in series?
zysi [14]

It increases

Explanation:

A circuit's resistance increases as more resistors are added to it in series.

Series connection of circuit provides an additive increase in the resistance offered by the overall circuit.

A resistor is a device in a circuit that impedes the flow of electric current. It simply uses electric current.

Examples are bulbs, radio, television.

The more of these devices connected one to another in series, the more the resistance increases.

learn more;

Voltage brainly.com/question/6949231

#learnwithBrainly

8 0
3 years ago
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
 
Leto [7]

Answer:

w

Explanation:

..

6 0
3 years ago
Read 2 more answers
Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 tim
Oksi-84 [34.3K]

Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity \rho.

The cross-sectional area of the of a wire is given by;

A=\pi\frac{d^2}{4}................(1)

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

\rho=\frac{RA}{l}..................(2)

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)

Given;

l_1=1.35l_2

\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)

We then use equation (1) to fine the ratio of the area A_1 to A_2

bearing in mind that d_1=0.91d_2

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

\frac{R_1}{R_2}= 1.35*0.8281=1.117935

4 0
3 years ago
Read 2 more answers
A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces
schepotkina [342]

Explanation:

(a)  Formula to calculate volume of the submerged wooden block is as follows.

            V_{sub} = l \times w \times d

It is given data of the wooden block is as follows.

          depth = 7.96 cm,       length (l) = 6 cm

         width (w) = 4 cm,      

So, we will calculate the volume of the submerged wooden block as follows.

           V_{sub} = l \times w \times d

                       = 6 \times 6 \times 7.96

                       = 286.56  cm^{3}

Hence,  the submerged volume of the block is 286.56  cm^{3}.

(b)   Expression for the buoyant force acting on the wooden block is as follows.

            F_{B} = \rho_{w} g V_{sub}

And, expression for the force of gravity of the wooden block is as follows.

            F_{g} = m_{b}g

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

                 F_{g} = F_{B}

Hence, mass of the wooden block will be calculated as follows.

         F_{g} = F_{B}

       m_{b}g = \rho_{w}gV_{sub}

          m_{b} = \rho_{w}V_{sub}

                      = 997 kg/m^{3} \times 286.56 cm^{3}

                      = 997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}

                      = 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c)   Expression for the density of the block is as follows.

             \rho_{b} = \frac{m_{b}}{V_{b}}

Now, expression for the total volume of the wooden block is as follows.

             V_{b} = l \times w \times h

Hence, density of the given block is as follows.

              \rho_{b} = \frac{m_{b}}{V_{b}}

                         = \frac{m_{b}}{lwh}

                         = \frac{0.02857 kg}{4 \times 4 \times 15}

                         = 1.19 \times 10^{-4} kg/cm^{3}

Therefore, density of the given block is 1.19 \times 10^{-4} kg/cm^{3}.

3 0
3 years ago
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