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blsea [12.9K]
3 years ago
9

Sjskskszpd no djdkzsnkakzx d

Physics
1 answer:
Mazyrski [523]3 years ago
4 0

Answer:

jadjsakdjadkjadhfmafkj

Explanation:

:)

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The climate of the earth throughout history has always _____.
Vlad [161]

Answer:

The climate of the earth throughout history has always been <em><u>fluctuated between hot and cold periods. </u></em>

7 0
3 years ago
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The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh
iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window b=35 in.

height of window h=20 in.

standard atmospheric pressure P_{outside}=14.7 psi

Also P_{inside}=1.01P_{outside}

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

F_{net}=(P_{inside}-P_{outside})\cdot A

F_{net}=P_{outside}(1.01-1)\times 35\times 20

F_{net}=14.7\times 0.01\times 35\times 20

F_{net}=102.9\ pounds\approx 103\ pounds            

8 0
2 years ago
A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second.
Leviafan [203]
Assuming Earth's gravity, the formula for the flight of the particle is: 

<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>

<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>

<span>Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet. </span>
7 0
3 years ago
Read 2 more answers
The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water
sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
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