The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.
<h3>What's the expression of force in terms of pressure?</h3>
- Mathematically, force = pressure/area
- Total area of the two hemispheres = 4π×(0.43)²= 2.3 m²
- Total pressure on the hemispheres= 15 milibar (directed inward) + 940 milibar (atmospheric pressure) = 955 milibar
=955×100 N/m²= 9.55×10⁴ N/m²
- Force on the hemispheres= 9.55×10⁴/2.3 = 4.2×10⁴ N
<h3>What's the minimum number of horses required to get 4.2×10⁴ N of force, if each horse can pull with a force of 1450N?</h3>
No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29
Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.
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<h2>64 Julios
</h2>
Explicación:
La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;
m es la masa del cuerpo
v es la velocidad del objeto
Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.
Parámetros dados
masa del objeto m = 0,5 kg
Variación de velocidad = 16 m / s
Variación de la energía cinética = 1/2 * 0,5 * 16²
Variación de la energía cinética = 1/2 * 0,5 * 256
Variación de la energía cinética = 0,5 * 128
<em>Variación de la energía cinética = 64 Julios</em>
The answer is : the same of yourself. <span>When you look at yourself in a pocket mirror and then hold the mirror farther away, you see the same of yourself.</span>
Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:32.24 s
Explanation:
Given
mass of runner (m)=51.8 kg
Constant acceleration(a)=
Final velocity (v)=5.47 m/s
Time taken taken to reach 5.47 m/s
v=u+at
Distance traveled during this time is
So remaining distance left to travel with constant velocity=153.57 m
thus time 
Total time=28.07+4.17=32.24 s
