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krek1111 [17]
2 years ago
11

Has an atomic number of 1

Physics
2 answers:
ololo11 [35]2 years ago
7 0

Answer:

1 = H = Hydrogen

tankabanditka [31]2 years ago
6 0

Answer:

The element H (Hydrogen) has an atomic number of 1.

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b
mars1129 [50]

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

8 0
3 years ago
What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change
sammy [17]

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

6 0
3 years ago
Acid rain is an example of which type of chemical weathering?
tino4ka555 [31]
The answer is Carbonic acid
6 0
3 years ago
Read 2 more answers
What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra
Tatiana [17]

Answer:

The centripetal acceleration of the car is 8\ m/s^2.

Explanation:

Let the mass of the car, m=10^3\ kg

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(20\ m/s)^2}{50\ m}

a=8\ m/s^2

So, the centripetal acceleration of the car is 8\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
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