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Elza [17]
3 years ago
14

Which of the following technologies can help you check your heart rate on the go?

Physics
2 answers:
Phoenix [80]3 years ago
7 0
The RHR is one of them.......
kupik [55]3 years ago
7 0
Your answer would be wrist monitor
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How does an object's position and velocity change as the object accelerates?”
taurus [48]

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

7 0
3 years ago
Understanding that if we say something unkind to someone else his or her
OlgaM077 [116]

Answer: Moral

Explanation:

6 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
1. Who first published the classification of the elements that is the basis of our periodic table today?
Gre4nikov [31]
That particular group of elements is reffered to as the "Noble Gasses"--a title that comes from the fact that these gases are very "secure" and don't mix well with other elements.
4 0
3 years ago
A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig
laila [671]

Answer;

C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation;

-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.

-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source

6 0
3 years ago
Read 3 more answers
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