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andrey2020 [161]

3 years ago

12

Match the definition with the correct term: the point at which the sun is as far north or as far south of the equator as possibl

e A. Perihelion B. Solstice C. Equinox D. Aphelion

1 answer:

White raven [17]3 years ago

5 0

Those are the two points on the map of the stars called
the "solstices" ... one in June and one in December.

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An athlete wants to refill an old punching bag used for training, as shown in the picture. When refilled, the bag will rip if it

In-s [12.5K]

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states action and reaction are equal and opposite. That is the force applied to an object is equal in magnitude to force experienced by the object but in opposite direction.

From the given question, when the bag exert a certain on the athlete, the athlete also exerts similar force to the bag but in opposite direction.

Thus, the complete sentence is as follows;

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

#SPJ1

5 0

2 years ago

In a photovoltaic system, an inverter is required to:.

Ivanshal [37]

Answer: convert the DC power produced by the modules into alternating current that can power lights, motors, and other loads.

Explanation:

5 0

3 years ago

A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?

Dmitriy789 [7]

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

6 0

3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

Please help u guys acellus sucks

34kurt

Answer:

6.0cm

Explanation:

Given

focal length = 15.0cm

object distance = 10.0cm

Required

Image distance v

Using the formula

1/f = 1/u + 1/v

1/15 = 1/10+1/v

1/v = 1/15 + 1/10

1/v = 2+3/30

1/v = 5/30

v = 30/5

v = 6.0cm

Hence the image distance is 6.0cm

7 0

3 years ago