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2 years ago

A ball is thrown at an angle of 38° to the horizontal. What happens to the

Physics

1 answer:

finlep [7]2 years ago

8 0

Answer:

Vy = V0 sin 38 where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g = Vy / g final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g final velocity = -Vy

time upwards = time downwards (conservation laws)

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Free electrons may be transferred between bodies by:

Tasya [4]

<span>Free electrons may be transferred between bodies by "Contact Action"

When positively charged body comes close to a negatively charged body, then electron flows from positively to negatively charged body

In short, Your Answer would be Option B

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2 years ago

Read 2 more answers

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

labwork [276]

The answer is already in the blank for, its was greater

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3 years ago

. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum

Tju [1.3M]

Explanation:

120+60=180

120-60=60

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2 years ago

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

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2 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$