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3 years ago

Explain some similarities and differences between: Magnetism and gravity.

Physics

2 answers:

AlekseyPX3 years ago

7 0

Answer:

Gravity attracts thing towards iself and magnet attracts mettalic things towards it self..

The force of gravity is equal everywhere but in magnets force of attraction is most near poles.

Gravity attracts everything towards it self but magnet only attracts mettalic objects.

Magnet could be temporarily made by passing electricity in any soft iron but gravity is natural can't be made..

svlad2 [7]3 years ago

4 0

Answer:

Gravity always attracts. Magnetism either attracts or repel. Gravity reacts to mass or space. Magnetism reacts to motion

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lidiya [134]

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3 years ago

A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

lions [1.4K]

Answer:

Approximately $4.61\times 10^{3}\; {\rm N}$ upwards (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ .)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) and
Normal force from the floor (upwards.)

Let $(\text{normal force})$ denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

$\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ .

Let $m$ denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be $(\text{weight}) = m\, g$ .

Let $a$ denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be $(\text{net force}) = m\, a$ .

Rearrange $\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ to obtain an expression for the magnitude of the normal force on this astronaut:

$\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}$ .

3 0

2 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

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3 years ago

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yarga [219]

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3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago