If a hockey stick exerts a force of 40 N on a 0.5-kg puck, what will the

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

PLEASE HELP!! THIS IS DUE RN!!

Oliga [24]

Answer:

know on car wright now

Explanation:

4 0

3 years ago

Ship A is 32 miles north of ship B and is sailing due south at 16 mph. Ship B is sailing due east at 12 mph. At what rate is the

Zielflug [23.3K]

Answer:

$\dfrac{dz}{dt} =-5.6\ mile/h$

Explanation:

distance between ship A and B = 32 mile

Ship A velocity in south, dx/dt = -16 mph

Ship B is sailing toward east with speed, dy/st = 12 mph

time = 1 hour

rate of change of distance between them = ?

x be the distance travel after t time

X = 32 + x

Let distance between them be z

now, using Pythagoras theorem to calculate distance between ships after 1 hours

z² = x² + y²

z² = (32 + x)² + 12²

z² = (32 - 16)² + 12²

z = √400

z = 20 miles

now, calculation of rate of change of distnace

z² = (32 + x)² + y²

differentiating both side w.r.t. time

$2 z \dfrac{dz}{dt} = 2(32+x)\dfrac{dx}{dt} + 2 y\dfrac{dy}{dt}$

$z \dfrac{dz}{dt} =(32-16)\dfrac{dx}{dt} +y\dfrac{dy}{dt}$

$20\times \dfrac{dz}{dt} =16\times (-16) +12\times 12$

$\dfrac{dz}{dt} =\dfrac{-112}{20}$

$\dfrac{dz}{dt} =-5.6\ mile/h$

hence, the rate is the distance between them changing at the end of 1 hour is equal to $\dfrac{dz}{dt} =-5.6\ mile/h$

7 0

3 years ago

Which choice has prokaryotic cells?

dem82 [27]

Answer:

Lactobacillus

Explanation:

classified in the prokaryotic kingdom Monera.

6 0

2 years ago

A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas

Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is $D = 1.38 m$

Explanation:

From the question we are told that

The length of the string is $L = 1.6 \ m$

The mass of the ball is $m = 200 g = \frac{200}{1000} = 0.2 \ kg$

The height of ball is $h = 1.5 \ m$

Generally the work energy theorem can be mathematically represented as

$PE = KE$

Where PE is the loss in potential energy which is mathematically represented as

$PE =mgh$

Where h is the difference height of ball at A and at B which is mathematically represented as

$h = y_A - y_B$

So $PE =mg(y_A - y_B)$

While KE is the gain in kinetic energy which is mathematically represented as

$KE = \frac{1}{2 } (v_b ^2 - 0)$

Where $v_b$ is the velocity of the of the ball

Therefore we have from above that

$PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)$

Making $v_b$ the subject we have

$v_b = \sqrt{2g (y_A - y_B)}$

substituting values

$v_b = \sqrt{2g (1.5 - 0.40)}$

$v_b = 4.6 \ m/s$

Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have

$v_x = 4.6 m/s \ while \ v_y = 0 m/s$

The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion

$s = v_y t - \frac{1}{2} g t^2$

Where s is distance traveled vertically which given in the diagram as $s = -0.4$

The negative sign is because it is moving downward

Substituting values

$-0.4 = 0 -\frac{1}{2} * 9.8 * t^2$

solving for t we have

$t = 0.3 \ sec$

Now the distance traveled on the horizontal is mathematically evaluated as

$D = v_b * t$

$D = 4.6 * 0.3$

$D = 1.38 m$

8 0

3 years ago