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Verizon [17]
3 years ago
10

The paths of fragments of fireworks are called

Physics
1 answer:
spayn [35]3 years ago
3 0

Answer:

Debree

Explanation:

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What is the smallest unit that makes up matter
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The smallest unit that makes up matter is an atom!
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3 years ago
Read 2 more answers
Suppose you wanted to break a meter stick over your knee. The cross-section of a meter stick is rectangular. Will it be easier t
LenaWriter [7]

Answer:

It will be easier to break the meter rule with the long side against my knee.

Explanation:

To break the meter rule involves the principle of bending moment. The long side will require less force to generate the same amount of bending moment that will have to be generated to break the meter rule. The short side on the other hand will require more force to generate this mount of bending moment. This is because the shorter has a very small surface area, which concentrates the force on your knee. The pressure is then dissipated as more pressure to your knee. Th longer side has a lesser surface area so, most of the force is used in breaking the meter rule.  

6 0
3 years ago
when an object is placed near a concave mirror, at what position does it forms a magnified and erect image.​
alexira [117]

Answer:

Between the principal focus and the pole of the mirror

4 0
3 years ago
A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?
Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

8 0
2 years ago
A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0
3 years ago
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