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swat32
3 years ago
13

Help me please,

Physics
1 answer:
garik1379 [7]3 years ago
7 0

Option B

Explanation:

no distance was given only the acceleration due to the fact that it went up (10m/s/s)

s0 it is

0 m/s and 10m/s/s (option B)

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Vapor is usually made with water and it would be made with gas. 
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4 years ago
A force of 6N is exerted on a cart that is carrying a rock and produces an acceleration of 2 m/s2. If the cart has a mass of 1 k
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If you use the equation F=ma and re-arrange it to get m = F/a, you'll get the total mass. Then you have to subtract the mass of the cart from the total mass to get the mass of the rock. 

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel
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Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

\vec v_{ra} = 12 m/s North

so it will be

Let North direction is along Y axis and East direction is along X axis

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also it is given that speed of air is 6.7 m/s relative to ground

\vec v_a = 6.7 \hat i

now as we know by the concept of relative motion

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now by rearranging the terms

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now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

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7 0
3 years ago
Is it possible for a distance-versus-time graph to be
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No.  If time is the horizontal axis and distance is the vertical axis, then
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5 0
4 years ago
Read 2 more answers
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

while the vertical distance covered must be

d_y = 6 m

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d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

The vertical distance covered instead is

d_y = \frac{1}{2}gt^2

where we omit the term ut since the initial vertical velocity is zero. From this equation,

t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
4 years ago
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