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ASHA 777 [7]
2 years ago
15

A boat has a porthole window, of

Physics
1 answer:
jekas [21]2 years ago
7 0

Answer:

F = 534.6[N]

Explanation:

We must find the pressure exerted by the water at the depth of the boat, by means of the following equation.

P=Ro*g*h

where:

Ro = density of sea water = 1027 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = wáter Depth =  6.25 [m]

Now replacing:

P=1027*9.81*6.25\\P=62967.93[Pa]

The net force is:

F = P*A\\F = 62967.93*0.00849\\F = 534.6[N]

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Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .
vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

Ke=\frac{1}{2}Mass*Velocity^2

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0
3 years ago
Read 2 more answers
Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.
galina1969 [7]
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³

Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

Answer: 8.06 in³=1.321 x 10⁻⁴ m³
8 0
3 years ago
What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio
pantera1 [17]
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),
 
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite. 

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite. 
3 0
2 years ago
What is a central therapeutic technique of psychoanalysis?
finlep [7]

These techniques are central to psychoanalytic therapy. They can be used alone or in combination with one another. Their purpose is to increase awareness and foster insight into the client's behavior and emotions

please rate and thanks me if this helped

8 0
3 years ago
The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in
N76 [4]

Answer:

28.2 m/s

Explanation:

The range of a projectile launched from the ground is given by:

d=\frac{v^2}{g}sin 2\theta

where

v is the initial speed

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

\theta is the angle at which the projectile is thrown

In this problem we have

d = 81.1 m is the range

\theta=45^{\circ} is the angle

Solving for v, we find the speed of the projectile:

v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s

7 0
3 years ago
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