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nataly862011 [7]
3 years ago
13

A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t

he string is 0.62 m, and the tension in the string when the ball is at the top of the circle is 4.0 N. What is v
Physics
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

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Is the color spectrum simply a small segment of the electromagnetic spectrum?
ch4aika [34]

Answer:

yup, u r correct

Explanation:

3 0
3 years ago
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel
Salsk061 [2.6K]

Answer:

If the particle is an electron E_y = 3.311 * 10^3 N/C

If the particle is a proton, E_y = 6.08 * 10^6 N/C

Explanation:

Initial speed at the origin, u = 3 * 10^6 m/s

\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, m_e = 9.1 * 10^{-31} kg

Mass of a proton, m_p = 1.67 * 10^{-27} kg

The electric field intensity along the positive y axis E_y, can be given by the formula:

E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\

If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C

If the particle is a proton:

E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 6.08 * 10^6 N/C

8 0
3 years ago
Two long, straight wires both carry current to the right, are parallel, and are 25 cm apart. Wire one carries a current of 2.0 A
Artemon [7]

Answer:

x = 7.14 meters

Explanation:

It is given that,

Current in wire 1, I_1=2\ A

Current in wire 2, I_2=5\ A  

Distance between parallel wires, r = 25 cm

Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

6 0
3 years ago
HI PLEASE HELP, I ONLY HAVE 10 MINUTES
Whitepunk [10]

Answer:

um ok

?????

i dont like you get rejected

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Explanation:

5 0
3 years ago
What is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 30 m/sec
disa [49]
To find this we should use the equation Ek=1/2mv^2 where Ek is the kinetic energy, m is the mass and v is the velocity. So by this Ek=1/2*1*30^2 which then makes Ek=450 joules
5 0
3 years ago
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