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3 years ago

13

Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a

4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?

2 answers:

Brilliant_brown [7]3 years ago

7 0

Answer:

4.0 kilogramas

Explanation:

gayaneshka [121]3 years ago

6 0

I use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop

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A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

7 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west

Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

$v_B=+5\ m/s$ is the velocity of Bill with respect to Amy (which is stationary)

$v_c=15\ m/s$ is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

$v_B=+5\ m/s$

Therefore, Carlos velocity in Bill's reference frame will be

$v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s$

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0

2 years ago

Is it possible to put a distance versus time graph to be universal go mine

xz_007 [3.2K]

Yes I'm pretty sure you can

3 0

3 years ago

Which of the following is not an example of osmosis? Multiple Choice

NNADVOKAT [17]

Answer:

Sugar in coffee dissolves.

Explanation:

In this case we have a solution, which is defined as separating what was bound in some way, by homogeneously mixing the molecules of a substance into a liquid. Therefore, it is the resulting homogeneous mixture after dissolving any substance in a liquid.

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3 years ago