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A solid conducting sphere is given a positive charge q. How is the charge q distributed in or on the sphere?

leva [86]

Answer:

Explanation:

the sphere is solid and conducting, so the charge is uniformly distributed over its volume.

7 0

3 years ago

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An athlete does one push-up. In the process, she moves half of her body weight, 250 newtons, a distance of 20 centimeters. This

s2008m [1.1K]

We know that an athlete moves half of her body weight 250 N a distance of 20 cm.
So: F = 250 N, d = 20 cm = 0.2 m
The work formula:
W = F * d
W = 250 N * 0.2 m
W = 50 J
Answer: Her work after one push-up is 50 J.

6 0

3 years ago

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

lozanna [386]

You can find

1) time to hit the ground
2) initial velocity
3) speed when it hits the ground

Equations

Vx = Vxo

x = Vx * t

Vy = Vyo + gt

Vyo = 0

Vy = gt

y = yo - Vyo - gt^2 / 2

=> yo - y = gt^2 / 2

1) time to hit the ground

=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

=> t = √1.631 s^2 = 1.28 s

2) initial velocity

Vxo = x / t = 6.5m / 1.28s = 5.08 m/s

3) speed when it hits the ground

Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s

V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2

=> V = √ (183.56 m^2 / s^2) = 13.55 m/s

7 0

3 years ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$

The energy of the system having mass 2m is,

$E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2$

For the two expressions mentioned above remember that the variables mean

m = mass

$\omega =$ Angular velocity

A = Amplitude

The energies of the two system are same then,

$E_m = E_{2m}$

$\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2$

$\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}$

Remember that

$k = m\omega^2 \rightarrow \omega^2 = k/m$

Replacing this value we have then

$\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}$

But the value of the mass was previously given, then

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}$

$\frac{A_1}{A_2} = 1$

Therefore the ratio of the oscillation amplitudes it is the same.

5 0

3 years ago

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

So

$a = 0.003* 6 =0.018\ m/s^2$

8 0

3 years ago