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Harman [31]
3 years ago
14

An electromagnetic wave with high frequency and high energy is A. Safe for humans. B. Is helpful to humans. C. Is helpful to hum

ans
Physics
1 answer:
Vilka [71]3 years ago
4 0
B and C are both correct choices.
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The process of mitosis makes body cells which are also called_______ cells.<br> NEED DONE ASAP!!!!!
lubasha [3.4K]

Answer:

daughters cells

Explanation:

when people refer to “cell division,” they mean mitosis, the process of making new body cells. Meiosis is the type of cell division that creates egg and sperm cells. Mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself.

4 0
2 years ago
Read 2 more answers
The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el
Harrizon [31]

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

8 0
3 years ago
How long will it take a plane to fly 1256km<br> if it travels 500km/hr?
expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

d is the distance

v is the velocity

From the question we have

t =  \frac{1256}{500}  \\  = 2.512

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0
2 years ago
a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an
11111nata11111 [884]
When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school
6 0
3 years ago
A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0
3 years ago
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