The distance in meters she would have moved before she begins to slow down is 11.25 m
<h3>
LINEAR MOTION</h3>
A straight line movement is known as linear motion
Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.
To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.
From definition of speed,
speed = distance / time
Make distance the subject of the formula
distance = speed x time
distance = 15 x 0.75
distance = 11.25m
Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m
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This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.
The process of ultrasonic images uses the "echo" property of the sound waves.
Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.
Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.
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The attached picture shows the process of ultrasonic imaging.
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps
The pressure drop in pascal is 3.824*10^4 Pascals.
To find the answer, we need to know about the Poiseuille's formula.
<h3>How to find the pressure drop in pascal?</h3>
- We have the Poiseuille's formula,

- where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
- By substituting values and rearranging we will get the pressure drop as,

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.
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