The splitting of an atom is called Blank Space _________

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field directed

Hatshy [7]

Answer:

0.114m

Explanation:

From the general expression for the radius of the proton's resulting orbit, we have

$r=\frac{mv}{qB}$

where q is is the charge of the proton $1.6*10^{-19}C$

m is the mass of the proton $1.67*10^{-27}kg$

B is the magnetic field $0.040T$

and v i the speed.

to determine the speed, we use the expression

Kinetic Energy= $qV$

$1/2mv^{2}=qV$

where <em>V </em>is the voltage value i.e 1.0kv

and v is the speed

Hence, from simple rearrangement we have the speed v to be

$v=\sqrt{\frac{2Vq}{m}} \\$

if we substitute value, we have

$v=\sqrt{\frac{2*1000*1.6*10^{-19} }{1.67*10^{-27}}} \\$

carrying out careful arithmetic we arrive at

$v=4.38*10^{5} m/s$ .

using the value for the speed in the expression for the radius of the orbit as stated earlier, we have

$r=\frac{1.67*10^{-27}*4.38*10^{5}}{1.6*10^{-19}*0.04} \\$

$r=0.114m$

7 0

3 years ago

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.95 m and whose unstretched length is 0.65 m. Next the

nexus9112 [7]

Answer:

Explanation:

spring constant of spring = mg / x

= .4 x 9.8 / ( .95 - .65 )

=13.07 N / m

energy stored in spring = 1/2 k x²

= .5 x 13.07 x ( 1.2 - .65 )²

= 1.976 J

Let it goes x m beyond its equilibrium position

Total energy at initial point

= 1.976 + 1/2 m v²

= 1.976 + .5 x .4 x 1.6²

= 2.488 J

energy at final point

= mgh + 1/2 k x²

.4 x 9.8 x ( .55 + x ) + .5 x 13.07 x² = 2.488

6.535 x² + 2.156 + 3.92 x = 2.488

6.535 x² + 3.92 x - .332 = 0

x = .075 m

7.5 cm

4 0

3 years ago

An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11

Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

8 0

3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 61 V is applied. What is the resistance of the wire?

Alisiya [41]

Resistance = (voltage) / (current)

For this piece of wire . . .

Resistance = (61 volts) / (6 Amperes)

Resistance = (61/6) (V/A)

<em>Resistance = (10 and 1/6) ohms</em>

Since you know the voltage and current, the length doesn't matter.

7 0

3 years ago

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of th

never [62]

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

$f=\dfrac{20}{3.1}$

f = 6.45 Hz

The frequency of oscillator is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

k is the spring constant

$k=4\pi^2f^2m$

$k=4\pi^2\times (6.45)^2\times 0.029$

k = 47.62 N/m

So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.

4 0

3 years ago

The splitting of an atom is called Blank Space __________.