A psychologist studies if internet uses causes isolation

A circular flat coil that has N turns, encloses an area A, and carries a current i, has its central axis parallel to a uniform m

gogolik [260]

Answer:

A. Zero

Explanation:

The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :

F = N * I * A * B * sinθ

Where θ is the angle between the normal of the enclosed area and the magnetic field.

Since the normal of the area is parallel to the magnetic field, θ = 0

Hence:

F = NIABsin0

F = 0 or Zero

3 0

2 years ago

Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass

Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0

3 years ago

Derive the expression for the range of the ball as a function of its height. To do this you should use the following steps. In t

AVprozaik [17]

Explanation:

Thanks for this....

its very helpful...

4 0

2 years ago

Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h

PolarNik [594]

The kinetic energy is $9\cdot 10^{40}J$ .

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

$m=5\cdot 10^{31} kg$ is its mass

$v=216,000 km/h$ is the speed

First, we convert the speed from km/h to m/s:

$v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s$

Therefore, the kinetic energy of the comet is

$K=\frac{1}{2}(5\cdot 10^{31})(60,000)^2=9\cdot 10^{40}J$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

5 0

3 years ago

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

2 years ago