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AnnZ [28]
3 years ago
10

Adam takes a bus on a school field trip. The bus route is split into the five legs listed in the table. Find the average velocit

y for each leg of the trip. Then arrange the legs of
the trip from highest velocity to lowest.
Leg Distance (km) Time (min)
А 18
9
B
25
15
С
24
8
D
48
12
E
15
7
leg A
leg B
leg c
leg D
leg E
Physics
1 answer:
Tema [17]3 years ago
7 0

Answer:

leg D → leg C → leg E → leg A →leg B

Explanation:

PLEASE VOTE ME BRAINLYIEST!!!!!!!!

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Summarize Newton's three laws of motion
Sedaia [141]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

7 0
3 years ago
PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
pav-90 [236]

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

3 0
2 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula
igomit [66]

Answer:

Angular acceleration of the disk will be \alpha =10.714rad/sec^2

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius r=\frac{d}{2}=\frac{0.3}{2}=0.15m

Moment of inertia of disk is given by I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2

Force is given by F=4 N

Torque is given as \tau =Fr=4\times 0.15=0.6N-m

We also know that torque is given by \tau =I\alpha

0.6=0.056\times \alpha

\alpha =10.714rad/sec^2

5 0
3 years ago
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

7 0
3 years ago
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