Anybody know this answer?

Which of these processes describes the effect Earth's atmosphere has on Earth's hydrosphere?

Ede4ka [16]

Warm, moist air increasing ocean temp

6 0

3 years ago

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

trasher [3.6K]

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$

7 0

2 years ago

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

dalvyx [7]

The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s

3 0

3 years ago

True or False: The arrows on a motion map should point in the directions of motion.

slavikrds [6]

Answer:

true,true,false

Explanation:

its false because if it is equal it would show an arrow pointing left and a 20 and the same for the right

3 0

2 years ago

A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

adell [148]

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

Given values are;

m1= 470 kg

x= 4m

m2= 75kg

Cm = center of mass

g= acceleration due to gravity= 9.82 m/s^2

The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)

= 9221.4Nm

τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm

τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

6 0

3 years ago