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Elodia [21]
3 years ago
8

For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o

f the time. If an object is dropped from the top of a tower 490 ft high and hits the ground in 7 ​sec, how far did it fall in the first 5 ​sec?

Physics
2 answers:
jasenka [17]3 years ago
5 0

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

stealth61 [152]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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Part A : A cylindrical water tank 10cm high and 520cm in diameter is filled with solution with a density of 0.75 g/cm^3
lisov135 [29]

Answer: 0.1066\ psi, 15.611\ kN

Explanation:

Given

the height of the tank is h=10\ cm

The diameter of the tank is d=520\ cm

Density of solution \rho=0.75\ g/cm^3\ or\ 750\ kg/m^3

(a) Water pressure at the bottom of the tank is

P=\rho gh\\P=750\times 9.8\times 0.1\\P=735\ Pa

1\ Pa=0.000145038\ psi

\Rightarrow P=735\ Pa\ or\ 0.1066\ psi

(b) \text{Average force on the bottom is the product of pressure and area of the base}

F_{avg}=735\times \pi \cdot (\frac{520}{200})^2\\\\F_{avg}=735\times 3.142\times 6.76\\F_{avg}=15,611.34\ N\ or\ 15.611\ kN

6 0
3 years ago
A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of
DENIUS [597]

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × 10^{-3}  kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 ×10^{-2} m

to find out

the speed of the mass

solution

we will apply here conservation energy  that is

K.E + P.E = Total energy  ..................1

so that Total energy = K.E max = P.E max

we know  amplitude so we find out first P.E max that is  

PE max = K.E + P.E  

(1/2)kA² = (1/2)mv² + (1/2)kx²  

kA^² =  mv²+ kx²

so here v²  will be

v²  = k(A² - x²) / m  

v = √[(k/m)×(A² - x²)]  ............2

here x = (1/2)A   so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × 10^{-3} )×(3/4×(38.9 ×10^{-2})²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s

6 0
3 years ago
A circular test track for cars has a circumference of 4.7 km. A car travels around the track from the southernmost point to the
Alika [10]
Well, for the distance traveled, the car goes from the northernmost point to the southernmost point. So, it travels half of the circle's circumference = 4.7/2 = 2.35 km.

For the displacement, by going from the northernmost point to the southernmost point, the car basically just travels the diameter of circle.

So, using the formula: Circumference = 2πr = <span>πd

Hence, the d = C/</span>π = 4.7/<span>π = 1.49605... = 1.5 km (2 significant figures)
Therefore, displacement = 1.5 km</span>
7 0
3 years ago
During a visit to the beach, you get in a small rubber raft and paddle out beyond the surf zone. Tiring, you stop and take a res
Monica [59]

Answer:

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

Explanation:

The movement in the two parts is very different, when the surf zone has passed it is in a deeper part of the water where the seabed does not rise much, therefore due to the movement of the waves there is an upward oscillatory movement and descending, in this movement there is no horizontal displacement.

When it is within the southern zone, there is a rapid rise of the sea floor, which generates a horizontal movement, having a traveling wave, therefore your movement is more complicated, you can have some oscillating movement on the axis and, but in addition to this you have a horizontal movement that reaches you towards the beach, forming a Traveling wave.

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

3 0
3 years ago
A 7.0kg skydiver is descending with a constant velocity
Vikentia [17]

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude mg, in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, R, in the upward direction

So the net force on the skydiver is:

F=mg-R

where

m = 7.0 kg is the mass

g=9.8 m/s^2

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

F=ma

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

a=0

Therefore the net force is zero:

F=0

And so, we have:

mg-R=0

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

R=mg=(7.0)(9.8)=68.6 N

6 0
4 years ago
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