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3 years ago

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A father racing his son has 1/2 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.4 m

/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

1 answer:

Assoli18 [71]3 years ago

5 0

answer:idk because idk how to work it out

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The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

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

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

Read 2 more answers

38. You are fishing and catch a fish with a mass of

adoni [48]

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

6 0

3 years ago

M = 30.3kg<br>M = 40.17kg 9<br>R = 0.5m<br>G = 6. 67x10^11<br>F ?

Lena [83]

Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

4 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

4 years ago