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Virty [35]
3 years ago
10

Which are ways to build self esteem?

Physics
1 answer:
Jlenok [28]3 years ago
8 0

Answer:

Start running and start to exercise

Explanation:

I dont really have an explanation but its true you need to start exercising and eat healthier if you want to build a self-esteem

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Describe three important ways we use the electromagnetic spectrum in our everyday lives.
OlgaM077 [116]

Answer:

We use X-rays to help the injured, Radiowaves to communicate or entertain, and Visible light to see.

5 0
3 years ago
Number 5 and 6 i need just give answer
sashaice [31]

Explanation:

voltage = current × resistance

5.

12 V = 4.2 A × resistance

resistance = 12 V / 4.2 A = 2.857142857... Ohm

FYI :

4.2 A would be a lot for a small electronic device like a CD player. that would be 12×4.2 = 50.4 Watt, and the CD player would get really hot.

6.

120 V = current × 12 Ohm

current = 120 V / 12 Ohm = 10 A

8 0
2 years ago
Which of the following traits might be considered an adaptation fora rabbit living in the Arctic?
puteri [66]
Where are the following traits ?
7 0
3 years ago
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The probability that a battery will last 10 hr or more is 0.8, and the probability that it will last 15 hr or more is 0.11. Give
podryga [215]
We are given with the data that says the probability that a battery will last 10 hr or more is 0.8 and the probability <span>that it will last 15 hr or more is 0.11. In this case, the probability that the battery lasts at least 10 hours and even 15 hrs more is 0.11 / 0.8 or equal to 13.75 percent.</span>
5 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
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