I need help on this and the first person who will answer this correctly gets BRANLIST

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0

3 years ago

Select all that Apple

egoroff_w [7]

B is the correct answer since acid corrode on metals such as carbon, steel, zinc and such.

6 0

4 years ago

Guyz... Its my first question in this app... Pls do answer

Basile [38]

Answer:

1.414

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1. So:

1 sin 45° = n sin 30°

n = sin 45° / sin 30°

n = 1.414

Round as needed.

6 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago

What relationship do you see between a star colour and temperature

Andrei [34K]

Answer:

Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is

5 0

3 years ago

I need help on this and the first person who will answer this correctly gets BRANLIST​

I need help on this and the first person who will answer this correctly gets BRANLIST