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2 years ago

5

A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replace

d with a 2.8kg mass.What is the stretch of the spring?

1 answer:

Anarel [89]2 years ago

8 0

2.0kg and the mass of sometimes makes it hard 2.8kg

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Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

3. The honeycomb-like appearance of this sandstone is a result of.

sladkih [1.3K]

Answer:

The answer is "Option C".

Explanation:

Wedging Freeze is generated by repeated freezing. Freeze wedging occurs, whenever the water is turned into ice as a result of the 9 percent expansion. When it freezes, cracks full of water are forced to separate further. and other options are incorrect, that can be described as follows:

In option A, It is a state where the element converts into a liquid, that's why it is not correct.
In option B, It is a reaction in which the bonds of water is divided into particular substance, that's why it is not correct.
In option D, It is a form of mechanical or physical rock weathering, that's it is not correct.

6 0

3 years ago

Read 2 more answers

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

alexdok [17]

Actual displacement that he required to move

$d_1 = 5.4 km$ towards North

Displacement that he moved due to snow is

$d_2 = 8.1 km$ at 47 degree North of East

now in vector component form we can say

$d_1 = 5.4 \hat j$

$d_2 = 8.1 cos47 \hat i + 8.1 sin47 \hat j$

$d_2 = 5.52 \hat i + 5.92 \hat j$

now the displacement that is more required to reach the destination is given as

$d = d_1 - d_2$

$d = 5.4\hat j - (5.52 \hat i + 5.92\hat j)$

$d = -5.52 \hat i - 0.52 \hat j$

so the magnitude of the displacement is given as

$d = \sqrt{5.52^2 + 0.52^2}$

$d = 5.54 km$

its direction is given as

$\theta = tan^{-1}\frac{0.52}{5.52} = 5.38 degree$

so it is 5.54 km towards 5.38 degree North of West.

4 0

2 years ago

Rearrange the kinetic energy equation so that velocity is the subject

mars1129 [50]

Answer: V = sqrt(2K.E/M)

Explanation:

K.E = 1 /2MV^2

MV^2 = 2 K.E

V^2 = 2K.E /M

V = sqrt(2K.E/M)

6 0

2 years ago