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Grace [21]
2 years ago
13

The best time to get my RESTING HEART RATE is right after I run.true or false ​

Physics
2 answers:
Savatey [412]2 years ago
8 0

Answer:

Your answer is: False

It is called a resting heart rate for a reason : )

Explanation:

Hope this helped : )

larisa [96]2 years ago
5 0

Answer: false

Your resting heart beat is when your heart beat is resting or calmed down. After you run, your heart is not so calmed. The answer is false.

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Bar graph

Explanation:

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2 years ago
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Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the “bottom
Sladkaya [172]

Answer:

1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.  

2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.  

3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.

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3 years ago
You just explored work and heat. Now test your understanding,
Lyrx [107]

Answer:

the hotter it gets, the liquid(mercury), expands more and more, and will rise up the tube to the correct line to read the tempature

5 0
2 years ago
"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much
san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, x=\dfrac{\lambda D}{d}      

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and \lambda is the wavelength of the light. Let x is the  number of bright fringes per unit width on the screen is given by :

x=\dfrac{n\lambda D}{d}

\lambda is the wavelength

n is the order

If n = 1,

x=\dfrac{\lambda D}{d}

So, the the number of bright fringes per unit width on the screen is \dfrac{\lambda D}{d}. Hence, the correct option is (B).

6 0
2 years ago
A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

8 0
3 years ago
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