Balance N3+H2 balance CaCo3+HCl balance K+Cl

(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if

Gwar [14]

Answer:

2324 J

Explanation:

The formula for work is:

$W=F*d$

where $F$ is the force applied, and $d$ is the distance moved, in this case $d=10.3m$

and we need to find $F.$

Since the crate is not moving up or down, we conclude that the normal force must be equal to the weight of the object:

$N=w$

where $N$ is the normal force and $w$ is the weight, which is: $w=mg$ , where g is the gravitational acceleration $g=9,81m/s^2$ and $m$ is the mass $m=46kg$ .

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Thus the normal force is:

$N=mg$

Now, the force due to the friction is defined as:

$f=\mu N=\mu mg$

where $\mu$ is the coefficient of friction, $\mu =0.5$

So, for the crate to move, the force applied must be equal to the frictional force:

$F=f\\F=\mu mg$

And now that we know the force we can calculate the work:

$W=F*d\\W=\mu mg*d$

substituting known values:

$W=(0.5)(46kg)(9.81m/s^2)(10.3)\\W=2324J$

5 0

3 years ago

Read 2 more answers

A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th

Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0

3 years ago

Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3

Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

$F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N$

$F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}$

The net force on q3 is the sum of these two forces:

$F_{net} = F_{13} + F_{23} = -1.92\times 10^{-5} + (-19.5\times10^{-6}) = 2.8\times 10^{-5} N$

Since both forces are directed towards left, their sign should be negative.

5 0

3 years ago

What are the advantages of having only one si unit for pressure???

Kazeer [188]

The units would be consistent around the world, allowing for easy comprehension of industrial diagrams and requirements as well as easier communication of engineers and scientists with one another.

6 0

3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0

3 years ago

Balance N3+H2balance CaCo3+HClbalance K+Cl​

Balance N3+H2balance CaCo3+HClbalance K+Cl