All categories

3 years ago

Which of these would represent the force of gravity

Physics

1 answer:

tigry1 [53]3 years ago

7 0

Missing figure: find it in attachment.

Answer:

Force D

Explanation:

In order to answer the question, let's keep in mind that the force of gravity on an object on Earth is the attractive force exerted by the Earth on the object; its direction is always downward (towards the Earth's centre), and its magnitude is given by

F = mg

where m is the mass of the object and g is the acceleration of gravity.

It follows immediately that in the figure, the force of gravity is the only force acting downward: therefore, force D.

The other forces are called:

Force A: thrust (it is the forward force generated by the engines)

Force B: lift (it is the upward produced by the aerodynamics of the wings)

Force C: air resistance (it is the backward force due to the friction between the air and the surface of the plane)

You might be interested in

What metric units are used to report the mass of an object

creativ13 [48]

Kilograms, hectograms, decagrams, grams, decigrams, centigrams, milligrams

8 0

4 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago

A voltmeter has an internal resistance of 10 000 Ω and is used to measure the voltage across a 47.0-Ω resistor that, without the

otez555 [7]

Answer:

b. 5.6 mA

Explanation:

As the voltmeter is connected to the resistor in parallel. The new resistance of the systems is

$\frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{10000} + \frac{1}{47} = 0.0214$

R = 1/0.0214 = 46.78 Ω

The voltage is

$U = r_2 I_1 = 47 * 1.2 = 56.4 V$

So the new current now of the system is

$I_2 = U/R = 56.4/46.78 = 1.2056 A$

So about 1.2056 - 1.2 = 0.0056 A or 5.6mA is drawn away.

7 0

4 years ago

One star has a magnitude of -1.5 and another star has a magnitude of +0.6. What is the ratio of brightness of these two stars?

In-s [12.5K]

Answer:

Explanation:

$The \ ratio \ of \ brightness \ of \ these \ two \ stars \ is:$

$\dfrac{b_{first \ star }}{b _{second \ star }}= 2.512^{(m_{(second \ star )} - (m_{first \ star } ))}$

$\dfrac{b_{first \ star }}{b _{second \ star }}= 2.512^{(0.6 - (-1.5 ))}$

$\dfrac{b_{first \ star }}{b _{second \ star }}= 2.512^{(2.1)}$

$\dfrac{b_{first \ star }}{b _{second \ star }}= 6.9$

$\mathbf{\dfrac{b_{first \ star }}{b _{second \ star }} \simeq 7}$

4 0

3 years ago

Ways we can reducing Surface Tension<br>

valina [46]

Surface tension can change with the change in a medium that is just above the layer of the liquid's surface.

Explanation:

Pouring any oil or oily compounds (such as kerosene) on the free surface of the water will reduce the surface tension.

in the atmosphere directly affects the surface tension of the liquid.

If we increase the temperature of the water, then there is a high possibility of the surface tension of the water getting reduced, due to the fact that the net force of attraction is decreased.

Mixing surfactants or emulsifiers into the water will decrease the surface tension.

If the water is subjected to electrification, then the surface tension will be reduced.

8 0

2 years ago

Read 2 more answers