Savanna regions developed during the Triassic period. is true
Answer:
D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force
Explanation:
For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.
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Hey There!</h2><h2>
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Answer:</h2><h2 /><h2>
![\huge\boxed{\text{V = 9.5 m/s}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Ctext%7BV%20%3D%209.5%20m%2Fs%7D%7D)
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<h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
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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)](https://tex.z-dn.net/?f=%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20W%5C%20%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5C%3B%28Base%29%5C%3A%28Height%29)
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](https://tex.z-dn.net/?f=W%5C%20%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5C%3A6%5C%3Ax%5C%3A30)
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](https://tex.z-dn.net/?f=%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20W%5Cquad%20%3D%5Cquad%20K.E%5C%5C%5C%5C%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20W%5Cquad%20%3D%5Cquad%20%5Cfrac%7B1%7D%7B2%7D%5C%20m%5C%20V%5E2%20%5C%5C%5C%5C%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20W%5Cquad%20%3D%5Cquad%20%5Cfrac%7B1%7D%7B2%7D%5C%202%5C%20%28V_f-V_i%29%5E2%5C%5C%5C%5C%7BV_i%5C%20is%5C%200%5C%20because%5C%20the%5C%20object%5C%20was%5C%20initially%5C%20at%5C%20rest%7D%5C%5C%5C%5C%20%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20W%5Cquad%5C%20%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5C%20x%5C%202%5C%20%28V_f-0%29%5E2%20%5C%5C%5C%5C%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%2090%5Cquad%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5C%20x%5C%202%5C%20%28V_f%29%5E2)
![\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20V_f%5Cquad%20%3D%5C%20%5Csqrt%7B%5Cfrac%7B2%5C%20%2890%29%5C%20%7D%7B2%7D%7D%5C%5C%5C%5C%7B%5CLongrightarrow%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cboxed%20%7BV_f%5Cquad%20%3D%5C%209.48%5C%20m%2Fs%7D)
![\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}](https://tex.z-dn.net/?f=%5Cboxed%7BThe%5C%20Velocity%5C%20of%5C%20the%5C%20Object%5C%20of%5C%20mass%5C%202kg%5C%20at%5C%206%5C%20meters%5C%20of%5C%20distance%5C%20was%5C%209.48%5C%20m%2Fs%7D)
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
The gravitational force will be one quarter.
The gravitational force between two objects is given by the formula
F=GMm/r^2
here, r is the distance between the objects.
Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.
Answer:
4.91 x 10⁻⁷ m
Explanation:
the applicable formula is
v = fλ
where
v = velocity (i.e speed) = given as 3.0 x 10⁸ m/s
f = frequency = given asw 6.11 x 10¹⁴
λ = wavelength
if we rearrange the equation and substitute the values given above,
v = fλ
λ = v/f
= 3.0 x 10⁸ / 6.11 x 10¹⁴
= 4.91 x 10⁻⁷ m