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Makovka662 [10]
3 years ago
15

A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo

ttom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?
Physics
1 answer:
Talja [164]3 years ago
5 0

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

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xxMikexx [17]

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

3 0
3 years ago
A river has a steady speed 0.500 m/s. A student swims upstream a distance of1.00 km and swims back to the starting point. a) If
Ivahew [28]

Answer:

a) 33.6 min

b) 13.9 min

c)  Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).


Explanation:


The problem deals with relative velocities.

  • Call Vr the speed of the river, which is equal to 0.500 m/s
  • Call Vs the speed of the student in still water, which is equal to 1.20 m/s
  • You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
  • And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.

Now, you can state the equations for each section:

  • distance = speed × time
  • upstream: distance = (Vs - Vr) × t₁ = 1,000 m
  • downstream: distance = (Vs + Vr) × t₂ = 1,000 m

Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:

  • (Vs - Vr) × t₁ = 1,000 m
  • (1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
  • (1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
  • total time = t₁ + t₂ = 1429s + 588s =  2,017s
  • Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min

Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s


  • time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min

Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.

6 0
4 years ago
Solve 305809 square root by division method​
Rus_ich [418]
You use long division.

Find the largest number whose square is less than or equal to the number in the leftmost group (55 < 30 < 66). Take this number as the divisor and the quotient with the number in the leftmost group as the dividend (30). Divide and get the remainder (5 in this case).

Bring down the next pair 58. Add the divisor with the quotient and enter it with a blank on its right. Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. In this case 105 × 5 = 525, so we choose the new digit as 5. Get the remainder.

Bring down the next pair 09. Add the divisor with the quotient and enter it with a blank on its right. Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. In this case 1103 × 3 = 3309, so we choose the new digit as 3. Get the remainder.

Put the decimal point.

End of division (Remainder is 0 and next digit after decimal is 0).

√305809 = 553
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what will be the effect on the acceleration due to gravity of the earth if it is compressed to a the size the moon?​
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Answer:

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