Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,

Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

5 0

2 years ago

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago

A car is traveling in a straight line at a constant speed. The car's engine is exerting a constant force on the car. Explain why

kherson [118]

Answer:

because its not going down a long hill instead its going on a leveled street

5 0

2 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

kati45 [8]

Explanation:

The given data is as follows.

radius (r) = 3.25 cm, $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

$a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

$a_{tangential} = \alpha \times r$

= $11.6 rad/s^{2} \times 3.25 cm$

= 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$ .

6 0

3 years ago

After today i might not be here no more

Norma-Jean [14]

Answer:

aw why? are you deleting the app for school?

6 0

2 years ago

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