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3 years ago

For communication to take place, there has to be:

Physics

1 answer:

sasho [114]3 years ago

3 0

Answer:

C. Sharing of the message

Explanation:

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When did you notice a greater elevation of blood pressure and pulse?

ch4aika [34]

You would notice a greater elevation of blood pressure and pulse when an individual experiences hypertension. It is also known as high blood pressure. When blood vessels are blocked or are constricted due to several reasons, the pressure of the blood would be increased since there would be smaller area for the blood to flow. Consequently, the pulse or the heart rate would increase as well due to the increase in pressure. Factors that can cause hypertension would be lack of exercise, stress, when hormones are not regulated. It can also be caused by another illness like diabetes, kidney disorder, obesity and many other.

6 0

3 years ago

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?

SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0

3 years ago

A bicyclist accelerates from a stop to a speed of 12m/s in 3 seconds.what is her acceleration

AVprozaik [17]

Answer:

4m/s2

Explanation:

acceleration=v-u/t...v=12m/s,u=0m/s,t=3sec

a=12-0/3

=4m/s2

8 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

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

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$