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viva [34]
3 years ago
15

HELP PLEASE 20 ;OINTS

Physics
1 answer:
77julia77 [94]3 years ago
7 0
I’m pretty sure it’s C.
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11. What is the name of earth's layer on which we live?<br><br>​
ollegr [7]

Answer:

the crust

Explanation:

8 0
3 years ago
you measure that it takes 0.75 seconds for a leaf to fall from a tree to the ground. the leaf experiences air resistance as it f
ella [17]
There's not enough information given in the problem to calculate that answer.

A leaf falling from a tree on Earth, a sheet of printer paper falling off the back
of a truck on Venus, and a steel ball sinking through a bucket of Scotch whiskey
on Mars, might all reach the bottom in 0.75 second.  The time it would take each
of them to fall the same distance through a vacuum in the same place would be
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you <em>did</em> know the acceleration of gravity in each place.  All you can say is that
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8 0
3 years ago
Convert the arc length of 6.86 radians into cycles.
Nezavi [6.7K]

Answer:

1.092 cycles

Explanation:

We need to convert the arc length of 6.86 radians into cycles.

We know that,

1 radian = 0.1592 cycles

To convert 6.86 radians to cycles, we use unitary method.

6.86 radian = (0.1592 × 6.86) cycles

= 1.092 cycles

Hence, there are 1.092 cycles in 6.86 radians. Hence, this is the required solution.

6 0
3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
A car enters the freeway with a speed of 6.4m/s and
Vsevolod [243]
Chicken nuggets , are your answer , eat 20 of them and then start work,
5 0
3 years ago
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