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Naddika [18.5K]

3 years ago

14

HELP ASAP‼️ Can someone please answer this question I’m confusion❔

1 answer:

klemol [59]3 years ago

3 0

the enclosed may help. earth rotes and you and i are fixed to it. air flows it's own path ...

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A patient is administered 20 mg of iodine-131. How much of this isotope will remain in the body after 40 days if the half-life f

igor_vitrenko [27]

The formula for half-life is:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where A is the amount of iodine-131 initially and after 40 days, t is time, h is half-life of the isotope. Let's plug in our values to the equation:

$A_{final}=20(\frac{1}{2})^{\frac{40}{8}=0.625g$

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.

4 0

2 years ago

What do you know about tides?

Neporo4naja [7]

Tides are the rise and fall of sea levels caused by the combined effects of the gravitational forces exerted by the Moon and the Sun, and the rotation of the Earth. Tide tables can be used to find the predicted times and amplitude (or "tidal range") of tides at any given locale.

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8 0

2 years ago

By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)

Fofino [41]

Gravity on the surface of sun is given as

$g = \frac{GM}{R^2}$

here we know that

$M = 1.98 \times 10^{30} kg$

$R = 6.95 \times 10^8 m$

now we will have

$g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}$

$g = 273.4 m/s^2$

now we need to find the ratio of weight on surface of sun and on surface of Earth

$R = \frac{mg_{sun}}{mg_{earth}}$

$R = \frac{273.4}{9.8} = 28 times$

so weight will increase by 28 times

6 0

2 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago

A handwashing sink may be used for: A. Handwashing and Washing Dishes B. Handwashing and Washing Ready-to-Eat Foods C. Handwashi

fiasKO [112]

C. Handwashing and sanitizing.

6 0

2 years ago

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