I think the awnser is Dark Matter
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Answer: 130 newtons</h3>
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Explanation:
We'll need the acceleration first.
- The initial speed (let's call that Vi) is 8.0 m/s
- The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
- This happens over a duration of t = 4.0 seconds
The acceleration is equal to the change in speed over change in time
a = acceleration
a = (change in speed)/(change in time)
a = (Vf - Vi)/(4 seconds)
a = (0 - 8.0)/4
a = -8/4
a = -2
The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.
Here's a further break down of Sam's speeds at the four points of interest
- At 0 seconds, he's going 8 m/s
- At the 1 second mark, he's slowing down to 8-2 = 6 m/s
- At the 2 second mark, he's now at 6-2 = 4 m/s
- At the 3 second mark, he's at 4-2 = 2 m/s
- Finally, at the 4 second mark, he's at 2-2 = 0 m/s
Next, we'll apply Newton's Second Law of motion
F = m*a
where,
- F = force applied
- m = mass
- a = acceleration
We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg
So the force applied must be:
F = m*a
F = 65*(-2)
F = -130 newtons
This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.
The magnitude of this force is |F| = |-130| = 130 newtons
B. If you press that into a calculator it comes up with 153.6. You then shift the decimal point 2 times forward and you end up getting 1.5 x 10^2 V.
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
