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stellarik [79]
3 years ago
8

A 40 kg bike traveling down the road has a negative acceleration. What is true of the force? A. It is positive B. It doesn't exi

st C. It is negative D. It is zero
Help me pls.
Physics
2 answers:
Greeley [361]3 years ago
4 0

Answer:

C. It is negative

Explanation:

Per Newton's second law, the net force is the mass times the acceleration:

∑F = ma

If the acceleration is negative, the net force is negative.

S_A_V [24]3 years ago
3 0

Answer:

C. It is negative

Explanation:

As we know that mass of the bike is given as

m = 40 kg

now we know by Newton's law of motion

F = ma

here we know that

m = mass of the bike

a = acceleration of the bike

so here we will have

F = 40 (a )

now if the acceleration of the bike is in negative direction

then we will get a negative force on it as per above formula

so correct answer will be

C. It is negative

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agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain  time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

_____________________________________________________

How do you know the acceleration at a certain time?

Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

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How do you know the  Displacement at a certain time?

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4 0
3 years ago
Please help on this one?
otez555 [7]

help me w mine and ill try to help with yours


3 0
3 years ago
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

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