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GuDViN [60]
2 years ago
15

jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a

mass of 11.0 kg, what is the centripetal force on miguel? round to the nearest whole number.
Physics
2 answers:
Morgarella [4.7K]2 years ago
6 0
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
Dimas [21]2 years ago
3 0

the answer is 225 n , got it right

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How much heat Q2Q2 is transferred to the skin by 25.0 gg of steam onto the skin? The heat of vaporization for steam is L=2.256×1
astraxan [27]

Answer:

56400Joules

Explanation:

The quantity of heat required is expressed as;

Q = mL

m is the mass = 25g = 0.025kg

L is the latent heat of vaporization for steam = 2.256×10^6J/kg

Substitute into the formula as shown;

Q = 0.025×2.256×10^6

Q = 56400Joules

Hence the quantity of hear required is 56400Joules

3 0
2 years ago
Which term do modern psychologists prefer to use in place of short-term memory?
Vsevolod [243]

Correct answer: D). Working Memory

The short term memory is the part of the memory system in which the information is stored for 30 seconds. However, it can be increased by rehearsal. Short term memory is also called active memory or the working memory. The working memory is the part of the cognitive system that is responsible for storing information temporarily for processing. It is important for reasoning.

Hence, the correct answer would be option D.




7 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
2 years ago
At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo
lukranit [14]

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

8 0
2 years ago
Which of the following statements cannot be supported by Kepler's laws of planetary motion?
gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio \frac{a^3}{T^2} (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0
2 years ago
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