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babymother [125]
4 years ago
5

Two bobsledders push a 232kg bobsled. The bobsled reaches an acceleration of 1.2x10^5 m/s^2. What is the force that the bobsledd

ers needed to reach this acceleration?
Physics
1 answer:
horrorfan [7]4 years ago
8 0

Force = (mass) x (acceleration)

Force = (232 kg) x (1.2 x 10⁵ m/s²)

<em>Force = 27,840,000 Newtons</em>

<em></em>

<u>Notes</u>:

-- If they could reach this acceleration and keep it up for a little while, the sled would reach the speed of light in about 40 minutes.  Their team would get the gold medal for sure.

-- This acceleration is a little over 12,000 G's. The sled and the pushers will fall apart long before they reach it.

-- The force required (2.784 x 10⁷ Newtons) is the same as about 3 million pounds, or roughly 1,564 tons of force by EACH pusher.

-- If either of them isn't pushing exactly straight and horizontal, or if there's any air resistance, or if there's any friction between the sled runners and the snow, then they'll need MORE force to reach that acceleration.

Lotsa luck !

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The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed
padilas [110]

Answer:

<em>Details in the explanation</em>

Explanation:

<u>Vertical Launch</u>

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

h = -16t^2 + 64t

<em>Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.</em>

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

h = -16(1)^2 + 64(1)=64-16=48\ ft

The object is at a height of 48 feet

At t=2 sec

h = -16(2)^2 + 64(2)=128-64=64\ ft

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

h = -16(3)^2 + 64(3)=192-144=48\ ft

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

h = -16(4)^2 + 64(4)=256-256=0\ ft

The object is at ground level and has returned to the launch point.

5 0
3 years ago
What is role of force on the speed of moving object? ​
aivan3 [116]

Explanation:

this is the ans hope it works

8 0
3 years ago
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Who knows how to do this
zlopas [31]

Answer:

1) 6

2) 2

3) 11

4) 38

5) 13

6) 14

7) 4

8) 41

Explanation:

3 0
3 years ago
A powerful storm attacks the coast, with winds blowing at 89 miles per hour. what is the velocity of these winds?
PtichkaEL [24]
Let's convert the speed of the winds into SI units.
We know that
1 miles = 1609 m
1 hour = 3600 s
So, the velocity written in m/s is
v=89  \frac{miles}{h} =89  \frac{miles}{h} \cdot  \frac{1609 m/miles}{3600 s/h}=39.8 m/s
6 0
4 years ago
Read 2 more answers
Feeling a bit better about not getting the Nobel Prize, Meitner drove home from the rink. If her 1500kg car accelerated from res
lyudmila [28]

Answer:

<em>The force exerted by the car engine was 3000 N</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F=m.a

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+a.t

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

The question describes how Meitner drove home taking her car from rest to a speed of 20 m/s in 10 seconds. This provides us the following data:

vf=20 m/s, v0=0 (rest), t = 10 seconds.

From the kinematics equation, we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

\displaystyle a=\frac{20\ m/s-0}{10\ s}=2\ m/s^2

The force exerted by the car engine was:

F=1500\ kg\cdot 2\ m/s^2

F=3000\ N

The force exerted by the car engine was 3000 N

5 0
4 years ago
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