PLS HELP!! IN QUIZ RN!

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

Todo sistema heterogêneo é bifásico

strojnjashka [21]

I don't understand the language.....

3 0

3 years ago

The default unit of rotational speed is 'radians per second' or 'rad/s'. If a someone sits at the edge of a 5m radius carousel a

Blizzard [7]

The rotational speed of the person is 0.4 rad/s.

<h3>Rotational speed (rad/s)</h3>

The rotational speed of the person in radian per second is calculated as follows;

v = ωr

where;

v is linear speed in m/s
r is radius in meters
ω is speed in rad/s

ω = v/r

ω = 2/5

ω = 0.4 rad/s

Thus, the rotational speed of the person is 0.4 rad/s.

Learn more about rotational speed here: brainly.com/question/6860269

7 0

2 years ago

If two objects at different temperatures are in contact with each other, what happens to their temperatures?

miv72 [106K]

In that case, heat energy flows from the warmer object to the cooler one.
As heat flows from one to the other, the temperature of the warmer object
falls, and the temperature of the cooler object rises. When the temperatures
are equal, the flow of heat energy from one to the other stops.

6 0

3 years ago