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2 years ago

In which diagram 1 or 2 would the person be experiencing winter? explain

Physics

1 answer:

pogonyaev2 years ago

6 0

Answer: Diagram 2

In diagram 1, the person in the northern hemisphere gets more sunlight compared to someone in the southern hemisphere. So diagram 1 has summertime in the northern hemisphere.

Conversely, diagram 2 has the southern hemisphere get more sunlight, so the northern hemisphere is experiencing winter.

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Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o

Ainat [17]

Answer:

Abbie’s acceleration is (1/2) Zak’s acceleration.

Explanation

1. Data:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. Formulae

Ac = ω² r

3. Solution:

a) Abbie:

Ac₁ = ω² r₁ = ω² (1m)

b) Zack:

Ac₂ = ω² r₂ = ω² (2m)

c) Divide Ac₁ / Ac₂

Ac₁ / Ac₂ = ω² (1m) / [ω² (2m) ] = 1/2

⇒ Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

5 0

2 years ago

We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

Arisa [49]

Answer:

$31.6\:\mathrm{m/s}$

Explanation:

The elastic potential energy of a spring is given by $Us=\frac{1}{2}kx^2$ , where $k$ is the spring constant of the spring and $x$ is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by $KE=\frac{1}{2}mv^2$ , where $m$ is the mass of the object and $v$ is the object's velocity.

Thus, we have:

$Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2$

Substituting given values, we get:

$\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}$

4 0

2 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How lo

Arlecino [84]

Answer:

t = 4.08 s

R = 40.8 m

Explanation:

The question is asking us to solve for the time of flight and the range of the rock.

Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:

Δx = v_0 t + 1/2at²

We have these known variables:

(v_0)_y = 0 m/s
a_y = -9.8 m/s²
Δx_y = -20 m

And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.

-20 = 0t + 1/2(-9.8)t²
-20 = 1/2(-9.8)t²
-20 = -4.9t²
t = 4.08 sec

The time it takes for the rock to reach the ground is 4.08 seconds.

Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.

List out known variables:

v_0 = 10 m/s
t = 4.08 s
a_x = 0 m/s

We are trying to solve for:

Δx_x = ?

By using the same equation, we can plug these known values into it and solve for Δx.

Δx = 10 * 4.08 + 1/2(0)(4.08)²
Δx = 10 * 4.08
Δx = 40.8 m

The rock lands 40.8 m from the base of the cliff.

7 0

2 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

$\Delta P_f = P_f - P_i$