Which of the following correctly shows the chain of energy transfers that create surface currents on the ocean?

1. Indicate whether these objects or atoms are positively,<br> negatively or neutrally charged.

vovangra [49]

Answer:

Neutrally charged!!!!!!!!!!!!!!!!!!!!!

Explanation:

6 0

3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

tangare [24]

The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back

4 0

3 years ago

Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa

-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

$\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}$

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have

$\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C$

Similarly for capacitor 2

$\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C$

Capacitance of the second capacitor = 2C

6 0

3 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t

Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

= $\dfrac{2\pi }{1.5}$

= 4.189 rad/s

Angular acceleration α = $\dfrac{\omega}{t}$

= $\dfrac{4.189}{40}$

= 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

v = 2πr/T

= $\dfrac{2\pi \times 6.2}{1.5}$

= 25.97 m/s

So centripetal acceleration.

a = $\dfrac{v^2}{r}$

= $\dfrac{25.97^2}{6.2}$

= 108.781 m/s^2

= 11.1 g

in combination with the gravitation acceleration.

$a_{total} = \sqrt{(11.1g)^2+g^2}$

$a_{total}= 11.145 g$

6 0

4 years ago