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2 years ago

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A skydiver falls toward the ground at a constant velocity. Which statement best applies Newton’s laws of motion to explain the s

kydiver’s motion? An upward force balances the downward force of gravity on the skydiver. The skydiver’s inertia is zero, which no longer allows for acceleration. An unbalanced force pulls the skydiver toward the ground. The mass of the skydiver is so great that there is no acceleration

2 answers:

vlabodo [156]2 years ago

8 0

Answer:

The answer is (a.) An upward force balances the downward force gravity on the skydiver

The skydiver is falling at a constant velocity because the upward force is balancing the downward force of gravity. According to Newton, the opposite force balance each other. This is stated in Newton's second law of motion.

Law Incorporation [45]2 years ago

7 0

Answer: A) an upward force balances the downward force of gravity on the skydiver

Explanation: I got it correct on my test on edge

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What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

2 years ago

The acceleration due to gravity is higher on juniper Than on earth. The mass and weight of rock on juniper compared to that on e

soldier1979 [14.2K]

Answer:

2.5 times higher then that on the Earth

Explanation:

Gravity is higher on Jupiter then on Earth because Jupiter is much bigger, because of it's mass compared to Earth the gravity on Jupiter is about 2.4 - 2.5 times higher then Earths surface gravity which means a rock on Jupiter would be around "2.4 - 2.5 times as heavier then it would be on Earth."

Hope this helps.

3 0

2 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

2 years ago

A thin lens with a focal length of 6.0 cm is used as a simple magnifier by (a) what angular magnification is obtainable with the

Serhud [2]

Answer:

4.167

4.83871 cm

Explanation:

u = Object distance

v = Image distance = 25 cm

f = Focal length = 6 cm

Angular magnification is given by

$m=\frac{25}{f}\\\Rightarrow m=\frac{25}{6}\\\Rightarrow m=4.167$

The angular magnification of the lens is 4.167

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{6}=\frac{1}{u}+\frac{1}{-25}\\\Rightarrow \frac{1}{6}+\frac{1}{25}=\frac{1}{u}\\\Rightarrow \frac{1}{u}=\frac{31}{150}\\\Rightarrow u=4.83871\ cm$

The closest distance by which the object can be examined is 4.83871 cm

5 0

2 years ago

A wave on the ocean surface with wavelength 44 m travels east at a speed of 18 m/s relative to the ocean floor. If, on this stre

inessss [21]

Answer:

A)t=<u>1.375s</u>

B)t=11s

Explanation:

for this problem we will assume that the east is positive while the west is negative, what we must do is find the relative speed between the wave and the powerboat, and then with the distance find the time for each case

ecuations

V=Vw-Vp (1)

V= relative speed

Vw= speed of wave

Vp=Speesd

t=X/V(2)

t=time

x=distance=44m

A) the powerboat moves to west

V=18-(-14)=32m/s

t=44/32=<u>1.375s</u>

B)the powerboat moves to east

V=18-14=4

t=44/4=<u>11s</u>

3 0

2 years ago