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Alexandra [31]
3 years ago
5

Which of the following is example of magnitude in a specific direction

Physics
1 answer:
k0ka [10]3 years ago
8 0
1. Velocity and 2. Acceleration
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So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.
bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at²   Subtract ViT from both sides

D - ViT = 0.5at²    Divide both sides by 0.5t²

\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }    I wrote this step out a little more to show how your fraction will cancel

\frac{D - ViT}{0.5t^{2} }= a    I like to flip these around so the single variable is on the right

a = \frac{D - ViT}{0.5t^{2} }

7 0
3 years ago
A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs
max2010maxim [7]

Well the one that is closer can see and hear more

3 0
3 years ago
An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

4 0
3 years ago
When do you turn your wheels sharply left before backing slowly in a parallel parking maneuver?
Alex_Xolod [135]

Answer:

What is C - when your front bumper is even with the front vehicles back bumper.

Explanation:

Good Luck

8 0
2 years ago
A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4
LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0
3 years ago
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