A ball is dropped from a height of 10 meters how long was the ball in the air and what is the speed when it hits the ground

A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser sp

uranmaximum [27]

Answer:

a) 351351.35m/s

b) 1.044*10^{-8}kg/C

Explanation:

a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

$F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}$

b) the mass-charge ratio is given by:

$\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}$

hope this helps!!

5 0

3 years ago

Having _________is the most obvious difference between a eukaryotic and a prokaryotic cell, but there are other differences as w

8_murik_8 [283]

Answer:

1. d. a nucleus

2. b. Membrane-bound organelles

3. c. ten time smaller

4. b. animal cell

5. c. Bacteria

Explanation:

Hope this helps... :)

7 0

3 years ago

An insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30°C. A valve is now opened, and argon is allowed to escape un

zheka24 [161]

Answer: The final mass of the tank is 2.46kg

Explanation: All shown in the attachment.

Assumptions:

i. Argon is treated as an ideal gas at the specified conditions.

ii. Isentropic relation of ideal gas applies at the given conditions.

4 0

3 years ago

A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in

otez555 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$a_{max} = 0.00246 \ m/s^2$

b

$k =722.2 \ N/m$

Explanation:

From the question we are told that

The amplitude is $A = 1.8 \ cm = 0.018 \ m$

The period is $T = 17 \ s$

The test weight is $W = 13 \ N$

Generally the radial acceleration is mathematically represented as

$a = w^2 r$

at maximum angular acceleration

$r = A$

So

$a_{max} = w^2 A$

Now $w$ is the angular velocity which is mathematically represented as

$w = \frac{2 * \pi }{T}$

Therefore

$a_{max} = [\frac{2 * \pi}{T} ]^2 * A$

substituting values

$a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018$

$a_{max} = 0.00246 \ m/s^2$

Generally this test weight is mathematically represented as

$W = k * A$

Where k is the spring constant

Therefore

$k = \frac{W}{A}$

substituting values

$k = \frac{13}{0.018}$

$k =722.2 \ N/m$

7 0

3 years ago

A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas

Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

$\Phi = BA Cos\theta$

Where

B= Magnetic Field

A = Area

$\theta =$ Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

$\theta =$ 0 then our expression can be written as

$\Phi = BA$