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harkovskaia [24]
3 years ago
13

For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

n axis through point A perpendicular to the page? Is it clockwise, or is it counterclockwise? Show your work and give correct units.
Physics
1 answer:
Phantasy [73]3 years ago
8 0

Torque is equal position vector times (r) times force vector (F).  Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

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Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b
RSB [31]
First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula 
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds
4 0
3 years ago
The time for a sound wave to travel between two people is 0.80 s,
sasho [114]

Answer:

Explanation:

Using the below formula

Speed of sound = ( distance between observers) *2/(total time taken)

Now putt the given values ,

time taken = 0.80 sec

distance = 256 m

hence

V of sound= 256*2/0.80

V of sound = 640 m/sec

4 0
3 years ago
A softball has a mass of 0.18 kg. What is the gravitational force on Earth due to the ball, and what is the Earth's resulting ac
alexgriva [62]

Answer:F=1.7802

Explanation:

Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is

F=ma where f-force,m-mass and a-acceleration due to gravity

So we can just substitute

F-?.m=.18 and a9.89

F=.18×9.89

F=1.7802N

6 0
2 years ago
In an application, Germanium is
Zigmanuir [339]

Answer:

produce electronics

Explanation:

The uses of Germanium are recorded beneath: Germanium's principle use is to deliver strong state hardware, semiconductors and fiber optic frameworks. As a phosphor in fluorescent lights.

7 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
2 years ago
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