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jasenka [17]
3 years ago
15

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

a velocity of 20 m/s. What was its acceleration?
Physics
2 answers:
podryga [215]3 years ago
7 0
I believe it is -1.11 m/s^2. I will let you know if its correct
Rom4ik [11]3 years ago
5 0
Your answer is -1.11m/s2. i know this cause i had it right on a test. hope this helps everyone else looking for the answer
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Astronauts are trained for take-off in a high-speed centrifuge of 4.7 m radius that spins in the horizontal plane.
Leokris [45]

Answer:

a) 1.94 \frac{rad}{s}

b) 9.12\frac{m}{s}

c) Towards the center of the centrifuge

Explanation:

a)

Becuse the centrifuge rotates in circular motion, there's an angular acceleration tha simulates high gravity accelerations

a_{rad}=\omega r^{2}

with r the radius and ω the amgular velocity, in or case a_rad=3.5g so:

3.5g=\omega r^{2} and g=9.8\frac{m}{s^{2}}

solving for ω:

\omega=\frac{3.5g}{r^2}=\frac{3.5*9.8}{4.2^2}

\omega = 1.94 \frac{rad}{s}

b) Linear speed (v) and angular speed are related by:

v=\omega r =(1.94)(4.7)

v= 9.12\frac{m}{s}

c) The apparent weigth is pointing towards the center of the circle, becuse angular acceleration is pointing in that direction.

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2 years ago
If you throw a baseball while riding a bike, the ball's speed relative to the ground is the sum of the bicycle's speed and the s
Ostrovityanka [42]
Take for example driving by with a cake in your hand, then dropping it while going 30 mph. It will not drop directly down, it will gradually go in the direction you were driving while falling.
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3 years ago
Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 2 m. What is your acceleration?
likoan [24]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s

time for the change = 2 minutes = 120 seconds

Acceleration = (8 m/s) / (120 seconds)

Acceleration = 0.067 m/s²

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A street light is on top of a 8 foot pole. Joe,
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8/4 = y/y-x

8y - 8x  = 4y

y = 2x

y = 2 x 4

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Hope this helps
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3 years ago
A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?
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<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
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