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Olin [163]
3 years ago
12

A heavier person will have a lower blood level due to a greater amount of ___ in their

Physics
1 answer:
jasenka [17]3 years ago
6 0

Answer:

Blood alcohol content (BAC) is a measure of the amount of alcohol that is present in the blood.

A heavier person will have a lower blood alcohol level due to a greater amount of water in their body. The more a person weighs, the more water he/she tends to have in his/her body. The water has a diluting effect on the alcohol so the BAC will be lower in comparison with a person who is slimmer  and has less water in the body.

Explanation:

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
Why is it sometimes difficult for scientists to identify a a fold or fault?
Sauron [17]

Folds and faults are difficult to identify because they occur in the interior of rocks and also due to the dense nature of the materials.

<h3>What are faults and folds?</h3>

Faults are lines of weakness are present in materials dues to uneven positioning of the particles of the material.

Folds occurs when infolds occur in materials.

Faults and folds usually occur in rocks.

Folds and faults are difficult to identify because they occur internally and also due to the dense nature of the materials.

Learn more about faults and folds at: brainly.com/question/14240712

4 0
2 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
Examples of the six simple machines
lakkis [162]
Car, computer,drill,air plane,television, vending machine

5 0
3 years ago
Read 2 more answers
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Hunter-Best [27]

Answer:

The mass of the cart is 150 kg.

Explanation:

Given that,

Mass of a boy, m₁ = 50 kg

Initial speed of boy, u₁ = 10 m/s

Initial speed of car, u₂ = 0 (at rest)

The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s

Let m₂ is the mass of the cart. Using the conservation of momentum as follows :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg

So, the mass of the cart is 150 kg.

3 0
2 years ago
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