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gladu [14]
3 years ago
15

A 74 kg firefighter slides, from rest, 4.9 m down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that t

he frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor?
Physics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer:

Her speed is 9.8 meter per second

Explanation:

Newton's second law states that acceleration (a) is related with force (F) by:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

Here the only force acting on the firefighter is the weight F=mg so (1) is:

mg=ma

Solving for a:

a=g

Now with the acceleration we can use the Galileo's kinematic equation:

Vf^{2}=Vo^{2}+2a\varDelta x (2)

With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

Vf^{2}=2a\varDelta x

Solving for Vf

Vf=\sqrt{2g\varDelta x}=\sqrt{2(9.81)(4.9)}

Vf=9.8\frac{m}{s}

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Is there any machine that is 100% efficient? why?why not
denis-greek [22]

Answer:

No, it's not there.

Explanation:

For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.

7 0
3 years ago
Why does a geostationary satellite must orbit around Earth's equator, rather than in some other orbit (such as around the poles)
LiRa [457]

Answer:

A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.

Explanation:

5 0
2 years ago
How much heat is absorbed by a 71g iron skillet when its temperature rises from 11oC to 29oC?
diamong [38]
The right formula to use for this calculation is the heat capacity formula, 
Heat absorbed, Q = MCT, Where 
M = Mass of the substance = 71
C = Specific heat capacity for iron = 0.450 J/gc
T = Change in temperature = 29 - 11 = 18
Q = 71 * 0.450 *18 = 575.10
The amount of heat absorbed by the iron skillet is 575 J.
5 0
3 years ago
The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

4 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
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