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2 years ago

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A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A

bug crawls along the line from the turntable’s center at 3.5 cm s relative to the turntable. Assume it is initially moving in the positive x direction. At the moment the bug gets to the edge, what are the x and y components of the velocity of the bug?

1 answer:

Bond [772]2 years ago

6 0

Answer:

$\mathbf{V_x = 3.25 \ cm/s}$

$\mathbf{V_y = 1.29\ cm/s}$

Explanation:

Given that:

The radius of the table r = 16 cm = 0.16 m

The angular velocity = 45 rpm

= $45 \times \dfrac{1}{60}(2 \pi)$

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

$t = \dfrac{r}{v}$

$t = \dfrac{0.16}{0.035}$

t = 4.571s

So the angle made by the radius r with the horizontal during the time the bug gets to the end is:

$\theta = \omega t$

$\theta = 4.712 \times 4.571$

$\theta = 21.54^0$

Now, the velocity components of the bug with respect to the table is:

$V_x = Vcos \theta$

$V_x = 0.035 \times cos (21.54^0)$

$V_x = 0.0325 \ m/s$

$\text {V_x = 3.25 \ cm/s}$ $\mathbf{V_x = 3.25 \ cm/s}$

Also, for the vertical component of the velocity $V_y$

$V_y = V sin \theta$

$V_y = 0.035 \times sin (21.54^0)$

$V_y = 0.0129\ m/s$

$\mathbf{V_y = 1.29\ cm/s}$

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The values of t are <u>4.643 second</u> for the function $H(t)=-16t^2+73t+6$

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We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula $H(t)=-16t^2+vt+s$

$H(t)=-16t^2+73t+6$

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

$-16t^2+73t+6=0$

We solve the equation using quadratic formula $x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values a=-16, b= 73, c=6

$t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643$

Learn more about the batter misses with the help of the given link:

brainly.com/question/19475098

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1 year ago

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Explanation:

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2 years ago

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I have a hot air balloon that has 18.0 g helium gas inside of it. if the pressure is 2.00 atm and the temperature is 297k, what

Anastaziya [24]

M = molar mass of the helium gas = 4.0 g/mol

m = mass of the gas given = 18.0 g

n = number of moles of the gas

number of moles of the gas is given as

n = m/M

n = 18.0/4.0

n = 4.5 moles

P = pressure = 2.00 atm = 2.00 x 101325 Pa = 202650 Pa

V = Volume of balloon = ?

T = temperature = 297 K

R = universal gas constant = 8.314

Using the ideal gas equation

P V = n R T

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3 years ago

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A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif

Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

$R=\frac{v^{2}sin(2\theta)}{g}$

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

$gR=v^{2}sin(2\theta)$

$\frac{gR}{v^{2}}=sin(2\theta)$

$sin^{-1}(\frac{gR}{v^{2}})=2\theta$

$\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}$

so now we can substitute the given data:

$\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}$

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

$tan \theta = \frac{h}{45.5m}$

so we can solve this for h, so we get:

$h=45.5m*tan(0.05614^{o})$

which yields:

$h=0.0445m$

or

$h=4.45cm$

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2 years ago

A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off toge

natka813 [3]

Answer: 909 m/s

Explanation:

Given

Mass of the bullet, m1 = 0.05 kg

Mass of the wooden block, m2 = 5 kg

Final velocities of the block and bullet, v = 9 m/s

Initial velocity of the bullet v1 = ? m/s

From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve

m1v1 = (m1 + m2) v, on substituting, we have

0.05 * v1 = (0.05 + 5) * 9

0.05 * v1 = 5.05 * 9

0.05 * v1 = 45.45

v1 = 45.45 / 0.05

v1 = 909 m/s

Thus, the original velocity of the bullet was 909 m/s

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3 years ago