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lianna [129]
3 years ago
5

Plz HELP! All of the following are possible reasons to complete strength training (anaerobic) exercises, except:

Physics
2 answers:
horrorfan [7]3 years ago
8 0

Answer:

to treat or prevent injuries

Explanation:

Illusion [34]3 years ago
4 0

Answer:i would say "to treat or prevent injuries"

Explanation: anaerobic exercise is: "high intensity movements preformed in a short period of time."

to treat or prevent injuries you would want to do Aerobic exercise, which is an easy exercise like walking.

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Which type of light is emitted from objects with the highest temperatures?.
Flura [38]
Incandescence is light at high temp
8 0
2 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n
hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

5 0
3 years ago
Which is correct help asap
Wittaler [7]

Answer:

b

Explanation:

5 0
3 years ago
What are the very small particles that make up all matter?
MariettaO [177]

Answer:

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8 0
3 years ago
Read 2 more answers
A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .
Rzqust [24]

Answer:

a)  ΔV = 2,118 10⁻⁸ m³   b)  ΔR= 0.0143 cm

Explanation:

a) For this part we use the concept of density

    ρ = m / V

As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats

    m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg

   

    V = m / ρ

    V = 0.56 / 3.52

    V = 0.159 cm3

We use the relation of the bulk module

    B = P / (Δv/V)

    ΔV = V P / B

    ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹

    ΔV = 2,118 10⁻⁸ m³

b) indicates that we approximate the diamond to a sphere

    V = 4/3 π R³

For this part let's look for the initial radius

    R₀ = ∛ ¾ V /π

    R₀ = ∛ (¾ 0.159 /π)

    R₀ = 0.3361 cm

Now we look for the final volume and with this the final radius

    V_{f} = V + ΔV

    V_{f} = 0.159 + 2.118 10⁻²

    V_{f} = 0.18018 cm3

    R_{f} = ∛ (¾ 0.18018 /π)

    R_{f} = 0.3504 cm

The radius increment is

    ΔR = R_{f} - R₀

    ΔR = 0.3504 - 0.3361

    ΔR= 0.0143 cm

4 0
3 years ago
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