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I am Lyosha [343]
3 years ago
11

A coin is dropped from a balcony 52.8m high. How long does it take the coin to land in

Physics
1 answer:
Lynna [10]3 years ago
8 0

Answer:

like couple of seconds

Explanation:

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which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi
JulsSmile [24]
<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>
7 0
3 years ago
Water near the poles would most likely be stored as
solniwko [45]
Hey there! 

Answer: Glaciers

Water near the poles would most likely be stored as glaciers. Glaciers are slow moving rivers that are a buildup of ice and snow. 

Thank you!

5 0
3 years ago
que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?​
ella [17]

Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

m = \dfrac{1}{ l^2} \cdot \dfrac{T}{4\cdot f^2} }

From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

8 0
3 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
4 years ago
Read 2 more answers
What is the frequency of a photon with an energy of 3.7 x10-18 j?
nata0808 [166]
Use the formula

E=h.lambda
3 0
3 years ago
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