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rjkz [21]
4 years ago
12

Someone help me pls I will give brainlist

Physics
2 answers:
stiks02 [169]4 years ago
8 0

Answer:

C and F

Explanation:

Because that’s how weather people measure tempature

(Brainliest?)

sergeinik [125]4 years ago
6 0

Answer:

the answer is point no. a °C , c. K and d. °F

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¿Qué entiendes por un fenómeno eléctrico?​
Inessa05 [86]

Answer:

English        

Electrical phenomena are commonplace and unusual events that can be observed and that illuminate the principles of the physics of electricity and are explained by them. Electrical phenomena are a somewhat arbitrary division of electromagnetic phenomena

Los fenómenos eléctricos son eventos comunes e inusuales que se pueden observar y que iluminan los principios de la física de la electricidad y son explicados por ellos. Los fenómenos eléctricos son una división algo arbitraria de los fenómenos electromagnéticos.

6 0
3 years ago
Which of the following is the correct SI unit to use in measuring the mass of a boulder
never [62]
Definitely not A or B. It really depends on the size of the boulder... It should be kilograms, unless it's a extremely huge boulder.
7 0
3 years ago
Read 2 more answers
The same 2000 kg truck formerly traveling at 32 m/s, plows into a
kvasek [131]

Answer: |F| = 1.28 x 10⁵ N

Explanation:

an impulse results in a change of momentum

FΔt = mΔv

F = m(vf - vi)/t

F = 2000(0 - 32) / 0.5

F = -128,000

|F| = 1.28 x 10⁵ N

5 0
3 years ago
A 0.155 kg arrow is shot upward
Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed  at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0
3 years ago
A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti
Morgarella [4.7K]

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

6 0
4 years ago
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