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Pie
3 years ago
15

a person exerts a force of 55 n on the end of a door 74 cm wide. what is the magnitude of the torque if the force is exerted?

Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer: Torque T = 40.7 Nm

Explanation:

Torque or turning effect is the rotational equivalent of linear force, i.e moment of force about a point.

Given that the force is applied perpendicular to the edge (angle 90°)

Torque can be represented by T;

T = FrSin∅ = Fr ( sin90 = 1)

T = force × distance

T = F × r

F = 55N

r = 74cm = 0.74m

T = 55N × 0.74m

T = 40.7 Nm

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You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
Paladinen [302]

Answer:

(A)

\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

Solving for t

\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

Now, let's put in numbers:

g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec

\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0

Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

Solving for D, we have two possible solutions:

D=54.71,\ D=26,651.11

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec

\displaystyle t_2=\frac{54.71}{345}=0.16\ sec

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0
3 years ago
The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.
Doss [256]

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

7 0
2 years ago
What type of wave is the highest on the electromagnetic spectrum?
san4es73 [151]
Gamma waves are on the far right
8 0
3 years ago
A plane comes in for a landing at a velocity of 80 meters per second west. As it touches down, it decelerates at a constant rate
azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

1) First find the time it takes the plane to stop using the equation for the acceleration:

a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

a is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

7 0
3 years ago
Question 14 (2 points)
sveticcg [70]

Answer:

after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

7 0
2 years ago
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