Cross-training employees is only necessary in the genetic science industry.<br> O True<br> O False

An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

so here we have

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0

3 years ago

A ray of light has a wavelength of

MArishka [77]

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, $\lambda=284\ nm$

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

$n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm$

So, the wavelength in vacuum is equal to 428.8 nm.

6 0

3 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

4 0

3 years ago

calculate how much charge passes through the LED in 1 hour if the current flowing in the circuit below is 350mA

Morgarella [4.7K]

Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

<h3>What is Current?</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = Q/t

Where Q is the charge and t is time elapsed.

Given the data in the question;

Time elapsed t = 1hr = 3600s

Current I = 350mA = 0.35A

Charge Q = ?

We substitute our given values into the expression above to determine the charge.

I = Q/t

Q = I × t

Q = 0.35A × 3600s

Q = 1260C

Therefore, given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

Learn more about current here: brainly.com/question/3192435

#SPJ1

3 0

2 years ago

The gravitational field at the Moon in N/kg due to the Earth is approximately (G = 6.67 × 10-11 N m2/kg2, the mass of the Earth

rosijanka [135]

Answer:

F = 2.69 10⁻³ m [ N]

Explanation:

This exercise asks to calculate the gravitational field of the Earth on the lunar surface, let's use the universal gravitation law

F = G m M / r²

where m is the mass of the body, M the mass of the Earth and r the distance between the Earth and the Moon

F = (G M / r²) m

F = (6.67 10⁻¹¹ 5.98 10²⁴ / (3.85 10⁸)² ) m

F = 2.69 10⁻³ m [ N]

This force is directed from the Moon towards the Earth, therefore it reduces the weight of the body

8 0

3 years ago

Cross-training employees is only necessary in the genetic science industry. O True O False