I will give brainliest if right

5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the

Sergio039 [100]

I think it is c because it was slowing down I don’t know for sure

5 0

3 years ago

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If a force of 10 N acts on an object and an additional force of 6 N acts on the object in the same direction. What will be the n

Schach [20]

Answer: The 1st one is B. I'm pretty sure the 2nd one is A. and the 3rd one is D.

Explanation: I hope this helped, although I'm not sure for the second one

4 0

2 years ago

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The electron cloud model describes which of the following?

GalinKa [24]

1. I think it's the trails left by an electron as it moves around the nucleus.

2. The atomic number is the number of protons so it is 8.

3. It's mass is lowered but it is still the same element.

3 0

3 years ago

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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$

4 0

3 years ago

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above

ryzh [129]

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v that is

v² = u² - 2gh ............1 for upward speed

v² = u² + 2gh ............2 for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9 = 26.76

v² = 8² + 2(9.8) 1.9 = 101.24

v = 5.173 m/s ..............3

v = 10.061 m/s ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173 = 2.827 m/s .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

6 0

3 years ago