What happens to Earth’s temperature as cloud cover increases?

dean is slowing down on his skateboards. he starts at a speed of 5.5 m/s and slows to 1.0 m/s over a time of 3.0 seconds. what i

vekshin1

Use one of the equations of accelerated motion; V2=V1 + at ...see attached

6 0

3 years ago

What would be the resulting condition if Earth's axis was perpendicular to the Sun?

blagie [28]

Answer:

I belive it would be B or D, but B seems more likely

Explanation:

3 0

3 years ago

Read 2 more answers

three masses are connected by a light string that passes over a frictionless pulley as shown. (a) what is there acceleration of

Ket [755]

The mass on the left has a downslope weight of

W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N

The mass on the right has a downslope weight of

W2 = 8kg * 9.8m/s² * sin35º = 45.0 N

The net is 25.3 N pulling downslope to the right.

(a) Therefore we need 25.3 N of friction force.

Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º

25.3N = µ * 92.3 N

µ = 0.274

(b) total mass is 11.5 kg, and the net force is 25.3 N, so

acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²

tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N

Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √

hope this helps. :)

3 0

4 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

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

So

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

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

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

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

$\Delta P_f = P_f - P_i$