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3 years ago

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A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

and the gravitational force of the Earth on the car compare at this point?A. N > mg B. N < mg C. N = mg

1 answer:

dybincka [34]3 years ago

4 0

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

$Normal force = Gravitational force - centripetal force$

At the foot of a round hill

$Normal Force = centripetal force + Gravitational force$

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9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

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3 years ago

Which branch of physics deals with the study of force, energy, and motion?

dexar [7]

The correct answer is Mechanics

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2 years ago

Read 2 more answers

A 0.43-kg object mass attached to a spring whose spring constant is 561 N/m executes simple harmonic motion. If its maximum spee

VLD [36.1K]

Answer:

0.22 m

Explanation:

m = 0.43 kg, K = 561 N/m

Vmax = 8 m/s

Let the amplitude of the oscillations be A.

The formula for the angular frequency of oscillation sis given by

$\omega = \sqrt{\frac{K}{m}}$

$\omega = \sqrt{\frac{561}{0.43}}$

ω = 36.1 rad/s

The formula for the maximum velocity is given by

Vmax = ω x A

A = Vmax / ω

A = 8 / 36.1 = 0.22 m

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2 years ago

HELP PLEASE!!!<br><br> a= ?<br><br> v=<br><br> r=

ratelena [41]

Answer:

$a=\frac{v^2}{r}$

$v=\sqrt{a*r}$

$r=\frac{v^2}{a}$

Explanation:

This is the formula for centripetal acceleration in terms of the tangential velocity (v) and the radius of the circular motion (r). The expression for the acceleration is already given, so simply type it as shown:

$a=\frac{v^2}{r}$

For the velocity (v) multiply by "r" both sides and then use the square root to solve for v:

$a*r=v^2\\v=\sqrt{a*r}$

For the radius multiply both sides by r and then divide both sides by the acceleration (a) in order to isolate r completely:

$a*r=v^2\\r=\frac{v^2}{a}$

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3 years ago

After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average

loris [4]

Answer:

0.4455 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

Total mass is

$m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg$

Here the spring constant is not given so let us assume it as $k=36000\ N/m$

Here, the forces are balanced

$mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m$

The springs are compressed by 0.4455 m

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3 years ago