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KIM
2 months ago
8

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

back and forth, the maximum angle that the string makes with the vertical is 34.0∘. Friction can be ignored. At the low point of the sphere's trajectory, what is the tension in the string?
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
5 0
0.833 N. To determine this, we first calculate the vertical distance from the highest position of the pendulum to the lowest point. This involves finding the height difference, which in this case is given by y = 1.2 - 1 = 0.2 m. As the pendulum moves downward, its potential energy is transformed into kinetic energy, following the conservation of energy principle.
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Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3465]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C represents the charge's magnitude

and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
3 months ago
A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const
Keith_Richards [3271]

a) -1.54 m/s^2

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

mg=m\frac{v^2}{r}where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top g=9.8 m/s^2

= loop radius

vBy solving for v,

r=1200 m

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived fromv_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

v_A=550 km/h =152.8 m/s

Given the plane's deceleration is consistent, we can obtain it using the following equation:

s=\pi r=\pi(1200)=3770 m

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:N=m\frac{v_B^2}{r}

v_B

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m

6 0
3 months ago
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