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KATRIN_1 [288]
4 years ago
13

Displacement is zero, even though it might have traveled a considerable _____

Physics
1 answer:
Brrunno [24]4 years ago
3 0
Displacement is a vector quantity .
So it will be zero if the traveling body comes to the starting point. but it might have traveled a considerable "distance".;-)
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Discuss potential behavioral concerns for people should they travel to Mars
Anna35 [415]

Answer:

The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...

6 0
3 years ago
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
3 years ago
A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from
svlad2 [7]

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

the hoop has mass M and radius R.

disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

3 0
4 years ago
Your average speed on the first half of a car trip is 69.0 km/h. How fast do you have to drive on the second half of the trip to
vova2212 [387]

Answer:

13 km/h

Explanation:

Average speed = distance/time

Let the total distance and total time taken for the whole trip be d km and t hours respectively

Average speed for the whole trip = 82 km/h

d = 82t

The distance covered in the first half = d1/2

Time taken = t/2

Average speed = 69 km/h

69 = d1/2 ÷ t/2

d1 = 69t

The distance covered in the second half = d2/2

Time taken = t/2

Let the average sly for the see half be A

A = d2/2 ÷ t/2

d2 = At

d = d1 + d2

82t = 69t + At

At = 82t - 69t

At = 13t

A = 13t/t = 13 km/h

3 0
4 years ago
A 6.0-kg object moving 5.0 m/s collides with and sticks to a 2.0-kg object. after the collision the composite object is moving 2
vredina [299]
  <span>m1v1 + m2v2 = (m1+m2)*(-vf) 
6*5 + 2*v2 = (6 + 2)*(-2) 
v2 = -16/2 = -8 m/s i</span>
5 0
4 years ago
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