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QveST [7]
3 years ago
9

If you are over 240 pounds you should only preform the high intensity exercises. True or False?

Physics
1 answer:
shutvik [7]3 years ago
7 0
False. You don’t want to over work yourself :) start at a level you are comfortable with and continue with making it more and more harder. EX: you can run around your block and you feel tired but refreshed. Do it everyday until you feel like you can go further
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How do spectroscopes help with studying of distance stars?
Svetradugi [14.3K]

Answer:

You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson

8 0
3 years ago
Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively, kept at a distance of 10 m from
NNADVOKAT [17]

Answer:

Answer

F=G×

d

2

m×M

m = 50 kg

M = 120 kg

Distance, d = 10 m

G=6.7×10

−11

Nm

2

/kg

2

F=6.67×10

−11

×

10

2

50×120

F=6.67×60×10

−11

F=4.02×10

9

N

6 0
3 years ago
Read 2 more answers
What is 4 differences between saturated unsaturated and supersaturated solutions
Dennis_Churaev [7]

Answer:

Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.

3 0
3 years ago
If during the submerged weighing procedure air bubbles were to adhere to the object, how would the experimental results be affec
Mazyrski [523]

Answer:

see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas

Explanation:

We will propose this exercise using Archimedes' principle, which establishes that the thrust on a body is equal to the volume of the desalted liquid.

          B = ρ g V

The weight of a submerged body is the net force between the weight and the thrust

          F_net = W - B

we can write the weight as a function of the density

          ρ_body = m / V

         m =  ρ_body V

         W = mg

         W =  ρ _body g V

we substitute

         F_net= ( ρ_body -  ρ _fluid) g V

In general this force is directed downwards, we can call this value the apparent weight of the body. This is the weight of the submerged body.

          W_aparente = ( ρ_body -  ρ _fluid) g V

If some air bubbles formed in this body, the net force of these bubbles is

         F_net ’= #_bubbles ( ρ_fluido -  ρ_air) g V’

this force is directed upwards

whereby the measured force is

         F = W_aparente - F_air  

           

As we can see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas

6 0
3 years ago
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t
Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

y=\frac{1}{2}at^2

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}

6 0
3 years ago
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