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Sloan [31]
3 years ago
15

A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to

Physics
1 answer:
juin [17]3 years ago
7 0
A= LxW
= 6.7x5.4
= 36.18 cm2
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When is the angular momentum of a system constant?
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I think it’s a but I’m not 100% sure
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3 years ago
A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp
sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So period of oscillation T=2\times 3.14\times \sqrt{\frac{3.42}{12}}

T=2\times 3.14\times 0.533=3.347sec

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

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3 years ago
In a maneuver, the jet comes in for a landing on solid ground with a speed of 115 m/s, and its acceleration can have a maximum m
V125BC [204]

Answer:

17.1130952381 s

No

Explanation:

t = Time taken

u = Initial velocity = 115 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.72 m/s² (negative as it is decelerating)

From the equations of motion

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-115}{-6.72}\\\Rightarrow t=17.1130952381\ s

The minimum time required to stop is 17.1130952381 s

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-115^2}{2\times -6.72}\\\Rightarrow s=984.00297619\ m

The distance that is required for the jet to stop is 0.98400297619 km which is greater than 0.8 km. So, the jet cannot land on a small tropical island airport.

4 0
3 years ago
Which of these devices does NOT contain a magnet or electromagnet?
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A light bulb don’t have magnet
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3 years ago
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Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
swat32

Point X lies on a horizontal line. We can intuitively say that the slope of the graph at point X is 0, therefore the acceleration at point X is 0m/s²

Point Y lies on a downward slanting line. To calculate the slope of that line, let's apply this equation:

m = (y₂-y₁)/(x₂-x₁)

m = slope, (x₁, y₁) and (x₂, y₂) correspond to the coordinates of the line's endpoints.

Given values:

(x₁, y₁) = (7, 5)

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Plug in and solve for m:

m = (0 - 5)/(12 - 7)

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The acceleration at point Y is -1m/s²

Choice A

5 0
3 years ago
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