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3 years ago

14

A straight bar magnet is initially 4 cm long, with the north pole on the right and the south pole on the left. if you cut the ma

gnet in half, the right half will

1 answer:

Greeley [361]3 years ago

6 0

The right half will be a new bar magnet of 2cm with north pole on the right side and south pole on teh left.

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Velocity is a vector quantity which has both magnitude and direction. ... Net force is also a vector quantity which has both mag

GuDViN [60]

When the object is moving in the elliptical orbit, it means that the direction of its acceleration should be towards the two foci (plural of focus) of the ellipse to keep the elliptical motion. As force according to the Newton's second law: F = ma, the net force must be in the direction of the acceleration. As far as the magnitude of net force is concerned, you can use Newton's gravitational law to find its magnitude.

7 0

4 years ago

Read 2 more answers

Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that

lawyer [7]

As we know by the first law of thermodynamics

$Q = \Delta U + W$

here we know that

Q = heat given to the system

$\Delta U = U_f - U_i$

W = work done by the system

now here we can say

$\Delta U = 180 - 78 = 102 J$

$W = 64 J$

now we can say that heat will be given as

$Q = 64 + 102 = 166 J$

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0

3 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago

A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c

guapka [62]

Answer:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

$(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )$

Second Person

$(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}$

The final velocities of the two people after the snowball is exchanged is:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

6 0

3 years ago

A 900 kg car travelling at 12 m/s due east collides with a 600 kg car travelling at 24 m/s due north. As a result of the collisi

katrin [286]

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

8 0

3 years ago