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3 years ago

A mailman performed 588J of work lifting a box 1.4 m. How much force did the mailman use?

Physics

1 answer:

lord [1]3 years ago

7 0

As we know that work done due a force F is given by formula

$W = F.d$

here F = applied force due to which work is done

d = displacement of object

now here we will have

W = 588 J

d = 1.4 m

now we can find force by above formula

$588 = F* 1.4$

$F =\frac{588}{1.4}$

$F = 420 N$

So applied force by milkman is F = 420 N

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Answer:

adding to or more vectors together . When displacement vectors are added, the result is a resultant displacement. But any two vectors can be added as long as they are the same vector quantity.

Explanation:

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4 years ago

if a football player does 39,000 J of work, how much power does the football player exert in 5 minutes?

elena-s [515]

Power = Work / Time

= 39000 / 5

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4 years ago

As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d

Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

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ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

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4 0

3 years ago

Two point charges are on the y axis. A 3.90-µC charge is located at y = 1.25 cm, and a -2.4-µC charge is located at y = −1.80 cm

ladessa [460]

Answer:

a) 1.6*10^6 V

b) 13.35*10^6 V

Explanation:

The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

q1 = 3.90µC = 3.90*10^-6 C

q2 = -2.4µC = -2.4*10^-6 C

r1 = 1.25 cm = 0.0125 m

r2 = -1.80 cm = -0.018 m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace all the parameters in the equation (1):

$V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V$

hence, the total electric potential is approximately 1.6*10^6 V

b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:

r1 = 0.0150m - 0.0125m = 0.0025m

r2= 0.015m + 0.018m = 0.033m

Then, you replace in the equation (1):

$V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V$

hence, for y = 1.50cm you obtain V = 13.35*10^6 V

4 0

3 years ago

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 27.2

garri49 [273]

Answer:

c=326.5177 J/kg.K

Specific heat is c=326.5177 J/kg.K

Explanation:

In order ti find the specific heat, we will proceed as follow:

Formula we are going to use is:

$Q=m*c*\Delta T$

Where:

Q is the heat energy added

m is the mass of sample

c is the specific heat

$\Delta T$ is the temperature Rise.

First we will find the mass:

Weight=m*g (g is gravitational acceleration=9.8 m/s^2)

$m=\frac{weight}{g} \\m=\frac{27.2}{9.8} \\m=2.7755 \ kg$

Rearranging above formula:

$c=\frac{Q}{m* \Delta T}$

$c=\frac{1.45*10^4}{2.7755*16}$

c=326.5177 J/kg.K

Specific heat is c=326.5177 J/kg.K

5 0

3 years ago