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san4es73 [151]
3 years ago
11

What is the avarage speed during kick 1

Physics
1 answer:
vova2212 [387]3 years ago
8 0
Is there answers to the problem ?
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PLEASE HELP!!! WILL MARK BRAINLIEST AND GIVE 5 STAR!!!!!!!!!!
Sladkaya [172]

Answer:

B. develop hypotheses that will be subjected to empirical test.

Explanation:

The rational method of inquiry is most useful in the early stages of science to develop hypotheses that will be subjected to empirical test.

7 0
3 years ago
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A conductor carrying a conventional current in the positive-x direction is in a magnetic field whose vector points in the positi
a_sh-v [17]

Answer:

Positive Z direction (out of screen)

Explanation:

Magnetic force is given by F = il \wedge B. A quick way to gauge the components is to put your left middle finger on the direction of the current, your index on the direction of the magnetic field, and the thumb gives you the answer you want.

4 0
3 years ago
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A hydraulic machine can be used to lift extremely heavy objects. Why is the fluid in the hydraulic machine a liquid rather than
Sauron [17]
Hi,

The force that acts on hydraulic machine is heavy therefore the content must be something that cannot be compressed by that kind of force, the gas can easily be compressed while a liquid is nearly impossible to.
4 0
3 years ago
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Find change of velocity for:<br> 0-1.7 s<br> 1.7-6s
tekilochka [14]
The integral of acceleration is velocity. The area under the curve is an integral. You can see the relation here. So just take the area under those time intervals. I’m assuming 1.7 is where it crosses the x-axis

For the first interval:
Base = 1.7
Height = -2
[1.7*(-2)]/2 = -1.7 cm/s

For the second interval:
There are two triangles and a rectangle here.
Base #1 = 0.3
Height #1 = 1

Base #2 = 1
Height #2 = 1

Length = 3
Width = 1

Apply the area formulas to get an answer of 3.65 cm/s
3 0
3 years ago
Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied
liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

5 0
3 years ago
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