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4 years ago

The current in a wire increases from 5 A to 10 A, what happens to its magnetic field?

Physics

1 answer:

sdas [7]4 years ago

6 0

Answer:

The first one is twice as strong

The second one is half as strong

The third one is the force is reversed

Explanation: just did it

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A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

5 0

3 years ago

The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate

lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

$T^{2} = a^{3}$ (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of $1.50x10^{8} Km$ . That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

$a = 2.98(1.50x10^{8}Km)$

$a = 4.47x10^{8}Km$

That distance will be expressed in terms of astronomical units:

$4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}$ ⇒ $2.98AU$

Finally, from equation 1 the period T can be isolated:

$T = \sqrt{a^{3}}$

$T = \sqrt{(2.98)^{3}}$

$T = \sqrt{26.463592}$

$T = 5.14AU$

Then, the period can be expressed in years:

$5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr$

$T = 5.14 yr$

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0

3 years ago

A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at

ivanzaharov [21]

(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

4 0

4 years ago

9. A car driver brakes gently. Her car slows down front --

sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0

4 years ago

All ecosystems need<br> and<br> in order to survive.

Triss [41]

Answer:

An ecosystem is a community of living things and their non-living environment, and may be as large as a desert or as small as a puddle. An ecosystem must contain producers, consumers, decomposers, and dead and inorganic matter. All ecosystems require energy from an external source – this is usually the sun.

Explanation:

4 0

3 years ago