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3 years ago

10

A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force e

xerted by the plane on the block is

2 answers:

WINSTONCH [101]3 years ago

7 0

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Colt1911 [192]3 years ago

4 0

Answer:

Mg Cos theta

Explanation:

Choosing a coordinate system parallel and perpendicular to the plane, since the acceleration perpendicular to the plane is 0, the net force perpendicular to the plane is also 0. Therefore, the magnitude of the normal force equals the magnitude of the y component of the weight.

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A soccer player kicks a rock horizontally off a 37 m high cliff into a pool of water. If the player hears the sound of the splas

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<h2>Answer:</h2>

<em><u>(a). 16.741 m/s</u></em>

<em><u>(b). 15.75 m/s</u></em>

<h2>Explanation:</h2>

In the question,

Height of the cliff, h = 37 m

Time taken to reach the sound to us = 2.92 s

Speed of the sound in air at room temperature = 343 m/s

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Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

$t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s$

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Splash is heard after = 2.92 s

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<u>Time taken by sound to travel the shortest distance along the hypotenuse of the triangle</u> thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

$Distance=343\times 0.172\\Distance=59.02\,m$

So,

In the triangle using the Pythagoras Theorem,

Horizontal distance traveled is,

$D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m$

So,

Speed of throwing of ball is given by,

$Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s$

<em><u>Therefore, the speed of the ball = 16.741 m/s.</u></em>

(b).

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Speed of sound = 331 m/s

So,

<u>Distance traveled by sound</u> is,

$Distance=331\times 0.172\\Distance=56.932\,m$

So,

Distance traveled in the horizontal by ball is,

$Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m$

So,

Speed of the ball thrown is given by,

$Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s$

<em><u>Therefore, the speed of the ball = 15.75 m/s.</u></em>

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Read 2 more answers

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

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