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3 years ago

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

1 answer:

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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True or false? Scientific theories lack experimental support

goldenfox [79]

Answer:

Hey there!

It is false. Theories can be repeatedly tested to give people an explanation of the world.

Let me know if this helps :)

8 0

3 years ago

A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.60 kg is attached to the sprin

trasher [3.6K]

(a) 1.08 J

The elastic potential energy stored in the block at any position x is given by

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the displacement relative to the equilibrium position

Here we have

k = 860 N/m

x = 5.00 cm = 0.05 m is the position of the block

Substituting, we find

$U=\frac{1}{2}(860 N/m)(0.05 m)^2=1.08 J$

(b) 1.16 m/s

The total mechanical energy of the spring-mass system is equal to the potential energy found at point (a), because there the system was at its maximum displacement, where the kinetic energy (because the speed is zero).

At the equilibrium position, the mechanical energy is sum of kinetic and potential energy

E = K + U

However, at equilibrium position x = 0, so U = 0. Therefore, the kinetic energy is equal to the total energy found at point (a)

$E=K= \frac{1}{2}mv^2 = 1.08 J$

where

m = 1.60 kg is the mass of the block

v is the speed

Solving for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.08 J)}{1.60 kg}}=1.16 m/s$

(c) 1.00 m/s

When the block is at position x = 2.50 cm, the mechanical energy is sum of kinetic and potential energy:

$E=K+U=\frac{1}{2}mv^2 + \frac{1}{2}kx^2$

where

E = 1.08 J is the total mechanical energy

m = 1.60 kg is the mass

v is the speed

k = 860 N/m

x = 2.50 cm = 0.025 m is the displacement

Solving for v, we find

$v = \sqrt{\frac{2E - kx^2}{m}}=\sqrt{\frac{2(1.08 J)-(860 N/m)(0.025 m)^2}{1.60 kg}}=1.00 m/s$

8 0

3 years ago

Which is more work, pushing with 115 N over 15m or lifting 20N over 10 m?

Nikitich [7]

Work = (force) x (distance)

You could look at the two cases, and see right away that
the first one has more force acting through more distance,
so it must be more work. But since I just gave you the formula
for Work, let's calculate the amount of it for both cases:

First case: Work = (115 N) x (15 m) = 1,725 joules

Second case: Work = (20 N) x (10 m) = 200 joules

The first case involves 8.625 times as much work as the second case.

3 0

3 years ago

Read 2 more answers

Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generat

Phantasy [73]

Answer:

Power = 9.75 ×10^8 $\frac{kgm^2}{s^3}$