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mariarad [96]
3 years ago
13

40 POINTS CORRECT ANSWER

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0
F=-kx
so multiple the two numbers together and put a negative sign in front
 F=-19.25
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A particle is traveling in the positive direction along an x axis, at a constant 5 m/s. Which of the following
kvasek [131]
The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.
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A duck swimming on the surface of a pond has an
Gemiola [76]

From the given information:

  • Taking the movement of the Duck in the North as the x-direction
  • The movement of the Duck in the East direction as the y-direction

However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}

The acceleration is:

\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) =  4 seconds.

Using the first equation of motion:

v^{\to} = u ^{\to} + a^{\to} t

Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

\mathbf{v^{\to} =0.87500   \hat x- 1.01629 \hat y}

Thus, the magnitude is:

\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}

\mathbf{v^{\to} =\sqrt{1.79848}}

\mathbf{v^{\to} =1.34 \ m/s}

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

brainly.com/question/17108011?referrer=searchResults

4 0
2 years ago
A 40.0 kg box is placed on a 2.00 m tall shelf. If the box falls off the shelf, what is its kinetic energy as it strikes the gro
RUDIKE [14]
M = 40 Kg , g=9.8 m/s² , h = 2 m

PE = m g h

PE = (40) (9.8) (2)

PE = 784 J

KE = PE

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½ v² = 19.6

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V = 6.26 m/s

KE = ½mv²

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3 years ago
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Stels [109]
Since 1m/s goes to 4m/s in 3 seconds,
then 4m/s goes to 8m/s in 6 seconds.
So your answer would be:
4m/s to 8m/s in 6 seconds
3 0
3 years ago
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