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1 year ago

10

Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel

y against girl 2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl 1 moves. ?

1 answer:

Arturiano [62]1 year ago

5 0

Answer:

$\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

Explanation:

The center of mass of the system (two girls) is constant, as the velocity of the center of mass of the system is also constant.

$\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

<u>The initial velocity of the system is zero, since both girls are at rest.</u> So the velocity of the total system at any point should be zero as well.

$0 = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\\\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

This is true, because there is no friction between the girls and the ground. Otherwise, the velocity of the center of mass wouldn't be constant.

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The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

Elza [17]

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

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2 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

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2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

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2 years ago

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

BigorU [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$f= -75 \ cm = - 0.75 \ m$

b

$P = -1.33 \ diopters$

Explanation:

From the question we are told that

The image distance is $d_i = -75 cm$

The value of the image is negative because it is on the same side with the corrective glasses

The object distance is $d_o = \infty$

The reason object distance is because the object father than it being picture by the eye

General focal length is mathematically represented as

$\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}$

substituting values

$\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}$

=> $f= -75 \ cm = - 0.75 \ m$

Generally the power of the corrective lens is mathematically represented as

$P = \frac{1}{f}$

substituting values

$P = \frac{1}{-0.75}$

$P = -1.33 \ diopters$

7 0

2 years ago

The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i

faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

5 0

2 years ago

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