Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by
where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is
where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:
where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,
So the closest answer among the option is -5020 N.
Learn more about acceleration and forces:
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Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock:
Pebble:
(b)Calculate the magnitude of the acceleration of each object when released.
Rock:
Pebble:
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em><em>:</em>
m1 will be equal to the mass of the Earth and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth
.
Newton's second law can be used to know the acceleration.
(2)
<em>Case for the pebble </em><em>:</em>