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3 years ago

9

Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin

g is my physics textbook is horrible and I don't understand a thing in there. Help would be appreciated!

1 answer:

aniked [119]3 years ago

8 0

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°). The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started. The displacement is therefore 0 m.

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2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0

DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 m $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

Kinetic energy is given 14 J

Kinetic energy is equal to $KE=\frac{1}{2}mv^2$

So $\frac{1}{2}\times 1.7\times v^2=14$

$v^2=16.47$

v = 4.05 m/sec

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

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3 years ago

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

3 years ago