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2 years ago

14

A boy and a girl are riding a merry-go-round which is turning at a constant rate. the boy is near the outer edge, while the girl

is closer to the center. who has the greater tangential acceleration?

2 answers:

Artyom0805 [142]2 years ago

8 0

If the ride is turning at a constant rate, then the acceleration is all centripetal, and zero tangential.

scZoUnD [109]2 years ago

7 0

I think that the girl has greater tangential acceleration because she is closer to the center and the acceleration is greater there.

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A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w

astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\$

So to find t that is the same time for all the motion

$t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s$

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

$Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}$

5 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

1 year ago

Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.

kramer

You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.

8 0

2 years ago

Read 2 more answers

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago

In order to get a tree stump out of the ground, chains are connected to two trucks. One truck pulls with a force of 600 N to the

Black_prince [1.1K]

Answer:

The net force on the stump is 1000 N.

Explanation:

Given that,

Force 1 acting on the truck, $F_1=600\ N$ (due north)

Force 2 acting on the truck, $F_2=800\ N$ (due west)

We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :

$F=\sqrt{F_1^2+F_2^2}$

$F=\sqrt{600^2+800^2}$

F = 1000 N

So, the net force on the stump is 1000 N. Hence, this is the required solution.

3 0

2 years ago