An uncharged body is a body that has equal magnitude of positive as well as negative charge on it. So as soon as some charged body is placed near it, the charged body induces negative charges on the uncharged body.
Answer:
Special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.
Explanation:
- A joist hanger also known as a beam hanger is a mechanical device which is used to fasten joists and rafters.
- The rafters are the carried members to beams and headers are the carrying members.
Thus, special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.
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We cannot consider something to be work unless <span>force applied causes movement of an object in the same direction as the applied force. </span>
Answer: 183.8Hz
Explanation:
Given that,
wavelength of sound (λ) = 1.85 m
frequency of the sound (F) = ?
Recall that the speed of sound (V) in air is a constant with a value of 340m/s
So, apply the formula
V = F λ
340 m/s = F x 1.85 m
F = 340m/s / 1.85m
F = 183.8Hz
Thus, the frequency of the sound that is
generated is 183.8 hertz
Explanation:
Given:
x₀ = 0 m
y₀ = 0 m
v₀ = 20 m/s
θ = 35°
aᵧ = -9.8 m/s²
1) Find t when y = 0.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
0 = 0 + (20 sin 35°) t + ½ (-9.8) t²
0 = t (20 sin 35° - 4.9 t)
t = 0, t = 2.34
The ball stays in the air 2.34 seconds.
2) Find y when vᵧ = 0.
vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)
0² = (20 sin 35)² + 2(-9.8) (y - 0)
y = 6.71 m
The ball reaches a maximum height of 6.71 meters.
3) Find x when y = 0.
x = x₀ + v₀ₓ t + ½ aₓ t²
x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²
x = 38.4 m
The ball lands 38.4 meters from Tom.
4) Find v when y = 0.
vₓ = aₓ t + v₀ₓ
vₓ = (0) (2.34) + 20 cos 35°
vₓ = 16.4 m/s
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8) (2.34) + 20 sin 35°
vᵧ = -11.5 m/s
v = √(vₓ² + vᵧ²)
v = √((16.4)² + (-11.5)²)
v = 20 m/s
The ball has a speed of 20 m/s just before it lands.
