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3 years ago

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Cardiorespiratory fitness improves the efficiency of the cardiovascular and the respiratory systems in delivering oxygen to the

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algol [13]3 years ago

4 0

Your answer should be T

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A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of

Nonamiya [84]

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

6 0

3 years ago

Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,

otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.

7 0

3 years ago

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

$n_2sinr=n_1sini$

$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

5 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

If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch

Slav-nsk [51]

<span>0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%</span>

8 0

3 years ago