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3 years ago

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A net force of Fnet acts on an object, causing the object to accelerate at a rat of 9m/s/s. what is the acceleration rate in a s

econd trial, if a new net force of 2Fnet acts on the same object?
a) 3m/s/s
b) 4.5m/s/s
c) 18m/s/s
d) 81m/s/s

1 answer:

Readme [11.4K]3 years ago

7 0

The new acceleration is c) 18m/s/s

Explanation:

Net force, mass and acceleration of an object are related by Newton's second law of motion:

$F=ma$

where

F is the net force on the object

m is its mass

a is its acceleration

In the first trial of this problem, a net force of $F_{net}$ is applied to the object, causing an acceleration of

$a=9 m/s^2$

Calling the mass of the object 'm', this means that

$F_{net} = ma = 9m$ [N] (1)

In the second trial, the force applied is $2F_{net}$ , so we have

$2F_{net} = ma'$

where a' is the new acceleration. Substituting (1) into the second equation, we find:

$2(9m) = ma' \rightarrow a' = 18 m/s^2$

This is because the acceleration is directly proportional to the force applied: therefore, if the force applied doubled, the acceleration doubles as well.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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A coin is placed on a vinyl stereo record that is making 33 1/3 revolutions per minute on a turntable. (a) In what direction is

mamaluj [8]

Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2

Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:

ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.

w=2*π*f

We know that the turntable is set to 33 1/3 rev/m so

the frequency 33.33/60=0.55 Hz

then w=2*π*0.55=3.45 rad/s

Finally the centripetal acceleration at differents radii results equal:

r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2

r=0.1 ac=3.45^2*0.1=1.20 m/s^2

r=0.14 ac=3.45^2*0.14=1.66 m/s^2

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3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

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3 years ago

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Hello,

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4 years ago

You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.

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sorry - late reply...just stumbled across tis...hope u can still use it :)

By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>

and do = distance to object = +8cm

Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
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3 years ago

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Answer:

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Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

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3 years ago