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Gnesinka [82]
3 years ago
13

In order for your workout to be safe and effective be sure you do not exceed your maximum working heart rate and you recover at?

Physics
2 answers:
Blababa [14]3 years ago
6 0
100 beats per minute
stira [4]3 years ago
6 0

It would need to be at 100 beats or less

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An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0
3 years ago
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (F_{t}), vertical gravitational force (F_{g}) and vertical normal force (F_{n}), due to the surface. Since there is no vertical movement, F_{g} and F_{n} cancels it out. So, for this block, net force is horizontal due to the rope F_{t}.

The block of mass m is hanging from the pulley, so there is the force of the rope (F_{t}) and the gravitational force (F_{g}). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

F_{r}=m.a

a=\frac{F_{r}}{m}

a=\frac{18.8}{3.6}

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

F_{t}-m.g=m.a

18.8-9.8m=5.22m

15.02m=18.8

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: F_{g}=m.g.

The weight for the hanging block is

F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0
3 years ago
Metals in group 1 on the periodic table most commonly form which type of ion
givi [52]

Answer:

Group 1 - the alkali metals. The Group 1 elements in the periodic table are known as the alkali metals.

Explanation:

7 0
3 years ago
What role does induction play when lightning strikes earth?
Dominik [7]
The role lightning plays in earth is when the earth is charged with positive protons, the lightning is the electron.
7 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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