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Romashka-Z-Leto [24]
2 years ago
11

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?​

Physics
1 answer:
Ksju [112]2 years ago
6 0

Answer:

Explanation:

T = 2π\sqrt{L/g}

(T / 2π)² = L/g

g = 4π²L/T²

g = 4π²(0.75000)/(1.7357)²

g = 9.82814766...

g = 9.8281 m/s²

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Which of the following happens to an object in uniform circular motion?
podryga [215]

Answer:

As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

5 0
4 years ago
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
rusak2 [61]
From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67° 
7 0
3 years ago
A convex mirror is placed to the right of an object. The image formed by the mirror will be a
kotegsom [21]

Answer:If you look at the image of the toy car in the mirror, it will appear to be the same ... However, there is a virtual focal point on the other side of the mirror if we follow them ... Concave mirrors, on the other hand, can have real images. ... Naturally, in concave mirror, the closer the image to the mirror, the bigger the image formed.

8 0
2 years ago
Read 2 more answers
Above
GaryK [48]
It had to be b Druidic
8 0
2 years ago
A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a w
pychu [463]

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

4 0
3 years ago
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