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4 years ago

Do Earth’s oceans gain or lose water, considering evaporation and precipitation together? How much?

1 answer:

7 0

Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.

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You are pushing an 80.0 N wheelbarrow as shown in (Figure 1). You lift upward on the handle of the wheelbarrow so that the only

AleksandrR [38]

Answer:

Wl = 1740 N

Explanation:

maximum lift weight unaided = force exerted (F) = 650 N

length of the wheelbarrow (L) = 1.4 m

weight of the wheelbarrow (w) = 80 N

distance of center of gravity of the wheel barrow from the wheel = 0.5 m

distance of center of gravity of the load from the wheel = 0.5 m

find the weight of the load (Wl)

from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive

ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0

(F x 1.4) = ((Wl x 0.5) + (w x 0.5)

Wl =

Wl = 1740 N

6 0

2 years ago

If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.

motikmotik

Answer:

Explanation:

8 0

3 years ago

It takes a wave traveling on a string 1.2 s to complete a full cycle. You measure the distance it takes to repeat itself to be 1

MArishka [77]

Answer:

1.5 m/s.

Explanation:

The Speed of a wave is the product of the wavelength and its frequency or it is the ratio of it wavelength and its Period. It can be expressed mathematically as

V = λf or λ/T .............................. Equation 1

Where V = speed of the wave travelling on the string, λ = wavelength of the wave, T = period of the wave, F = frequency of the wave.

Given: λ = 1.8 m ( distance it takes to repeat itself), T = 1.2 s (Time taken to complete a full cycle)

Substitute into equation 1

V = 1.8/1.2

V = 1.5 m/s.

Hence the wave is 1.5 m/s fast.

4 0

3 years ago

Calculating the acceleration of two people pulling a box with same force and angle.

Dmitriy789 [7]

Here as we can see there we be net horizontal force acting on it

$F_x = F_1cos20 + F_2cos20$

also we know that

$F_1 = F_2 = 80 N$

now we will have

$F_x = 2(80cos20)$

$F_x = 150.35 N$

now for net force of box we know

$F_x - F_f = ma$

$150.35 - 100 = ma$

$20a = 50.35$

$a = 2.52 m/s^2$

so acceleration of the box will be 2.52 m/s/s

7 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$