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3 years ago

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Which of the following is an example of benefits?

1 answer:

dangina [55]3 years ago

6 0

Insurance would be a example of benefits because you don’t receive insurance like you would salary, salary is a base thing you receive when you get a job while insurance you only get as a benefit with certain jobs

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AYUDA :( Calcular el trabajo desarrollado por la fuerza de fricción, si la fuerza "F" desplaza al bloque de una distancia de 40m

kogti [31]

Answer:

sorry can you translate

Explanation:

to english

4 0

3 years ago

During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz elec

erik [133]

Answer:

(a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

Explanation:

Given that,

Maximum potential = 4.00 mV

Distance = 0.300\ m

Frequency = 1.00 Hz

(a). We need to calculate the maximum electric field strength

Using formula of the potential difference

$\Delta V=Ed$

$E=\dfrac{\Delta V}{d}$

$E=\dfrac{4.00\times10^{-3}}{0.300}$

$E=0.0133\ V/m$

(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave

Using formula of the maximum magnetic field strength in the electromagnetic wave

$B=\dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{0.0133}{3\times10^{8}}$

$B=4.433\times10^{-11}\ T$

(c). We need to calculate the wavelength of the electromagnetic wave

Using formula of wavelength

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{1.00}$

$\lambda=3\times10^{8}\ m$

Hence, (a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

4 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

What is the angular momentum at a radius of 2 m with an object of 5 kg at a<br> velocity of 20 m/s?

ahrayia [7]

Answer:

200

Explanation:

angular momentum=mvr

=2×5×20

= 200kgm2/s

3 0

2 years ago

Read 2 more answers

Electrical current is the flow of ____________

Amanda [17]

Solution :

A) Electric current is the flow of electrons .

B) We know, by ohm's law :

V = I × R

Putting given values in above equation, we get :

V = 0.56 × 72 V

V = 40.32 V

Hence, this is the required solution.

3 0

3 years ago