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2 years ago

Jarin drank a lot of water before going to the doctor. What best describes what would happen to his blood pressure?

Physics

2 answers:

Nadya [2.5K]2 years ago

6 0

Answer:Lower

Explanation:

When Jarin drink a lot of water then his blood pressure will be lowers down .

As dehydration causes some of capillary beds to shut down and it put more pressure on capillaries and arteries causing blood pressure to increase.

sashaice [31]2 years ago

4 0

It lowers your blood pressure for suree

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Pls helppp. Is this right?

andreyandreev [35.5K]

Answer:

yes you are totally right

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3 years ago

Read 2 more answers

A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,

Zarrin [17]

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

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2 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

2 years ago

If a planet has a non-circular orbit around a star, as the planet moves closer towards a star its orbital speed ….

makkiz [27]

I think the answer you’re looking for is
D)
-I hope this helps! Enjoy the rest of your day

3 0

3 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]

1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor + (V_capacitor/RC) = (V_source/RC)

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

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

2) Charge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

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

2 years ago