Atmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa, which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.
<h2>
Hope it helps...</h2>
Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity
![v_{12} = v_1 - v_2](https://tex.z-dn.net/?f=v_%7B12%7D%20%3D%20v_1%20-%20v_2)
here
speed of first plane is 700 mi/h at 31.3 degree
![v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j](https://tex.z-dn.net/?f=v_1%20%3D%20700%20cos31.3%5Chat%20i%20%2B%20700%20sin31.3%5Chat%20j)
![v_1 = 598.12\hat i + 363.7\hat j](https://tex.z-dn.net/?f=v_1%20%3D%20598.12%5Chat%20i%20%2B%20363.7%5Chat%20j)
speed of second plane is 570 mi/h at 134 degree
![v_2 = 570 cos134 \hat i + 570 sin134 \hat j](https://tex.z-dn.net/?f=v_2%20%3D%20570%20cos134%20%5Chat%20i%20%2B%20570%20sin134%20%5Chat%20j)
![v_2 = -396\hat i + 410\hat j](https://tex.z-dn.net/?f=v_2%20%3D%20-396%5Chat%20i%20%2B%20410%5Chat%20j)
now the relative velocity is given as
![v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j](https://tex.z-dn.net/?f=v_%7B12%7D%20%3D%20%28598.12%20%2B%20396%29%5Chat%20i%20%2B%20%28363.7%20-%20410%29%5Chat%20j)
![v_{12} =994.12\hat i -46.3 \hat j](https://tex.z-dn.net/?f=v_%7B12%7D%20%3D994.12%5Chat%20i%20-46.3%20%5Chat%20j)
now the distance between them is given as
![d = v* t](https://tex.z-dn.net/?f=d%20%3D%20v%2A%20t)
![d = (994.12 \hat i - 46.3 \hat j)* 3](https://tex.z-dn.net/?f=d%20%3D%20%28994.12%20%5Chat%20i%20-%2046.3%20%5Chat%20j%29%2A%203)
![d = 2982.36\hat i - 138.9\hat j](https://tex.z-dn.net/?f=d%20%3D%202982.36%5Chat%20i%20-%20138.9%5Chat%20j%20)
so the magnitude of the distance is given as
![d = \sqrt{2982.36^2 + 138.9^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B2982.36%5E2%20%2B%20138.9%5E2%7D)
miles
so the distance between them is 2985.6 miles
First, let us assign the variables
y = 0.90 m, x= 15 m,
![h_{0}](https://tex.z-dn.net/?f=%20h_%7B0%7D%20)
= 2.80 m
s= required
The vertical component of the trajectory is in uniformly accelerated motion. The equation is:
![y= v_{0,y} t+ \frac{1}{2} a t^{2} + h_{0}](https://tex.z-dn.net/?f=y%3D%20v_%7B0%2Cy%7D%20t%2B%20%5Cfrac%7B1%7D%7B2%7D%20a%20t%5E%7B2%7D%20%2B%20h_%7B0%7D%20)
, while the horizontal component is
![x= v_{0,x} t](https://tex.z-dn.net/?f=x%3D%20v_%7B0%2Cx%7D%20t)
. Also,
![v_{0,y} =0](https://tex.z-dn.net/?f=%20v_%7B0%2Cy%7D%20%3D0)
since the object starts from rest (with respect to the downward motion). a is negative because it is moving downwards (a = -9.81 m/s^2). Substituting,
![0.9= 0+ \frac{1}{2} (-9.81m/ s^{2} ) ( \frac{x}{ v_{0,x} } )^{2} + h_{0}](https://tex.z-dn.net/?f=0.9%3D%200%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28-9.81m%2F%20s%5E%7B2%7D%20%29%20%28%20%5Cfrac%7Bx%7D%7B%20v_%7B0%2Cx%7D%20%7D%20%29%5E%7B2%7D%20%2B%20h_%7B0%7D%20)
![0.9= 0+ \frac{1}{2} (-9.81m/ s^{2} ) ( \frac{15}{ v_{0,x} } )^{2} + 2.5](https://tex.z-dn.net/?f=0.9%3D%200%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28-9.81m%2F%20s%5E%7B2%7D%20%29%20%28%20%5Cfrac%7B15%7D%7B%20v_%7B0%2Cx%7D%20%7D%20%29%5E%7B2%7D%20%2B%202.5)
The magnitude of v0, which is speed (s), is equal to 24.1 m/s
Answer:
<h2><em>
0.165Tesla</em></h2>
Explanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
<em>Hence the magnitude of the magnetic field is approximately 0.165Tesla</em>
Answer:
The force exerted by the muscle is 852.27N
Explanation:
Force F exerted by the muscle is expressed as a function of the torque τ and the effective perpendicular arm r.
F = τ/r ... (1)
Where τ = moment of inertia I × angular acceleration α
τ = Iα ... (2)
Substituting equation 2 into 1 to get F will give;
F = Iα/r
Given the following parameters
I = 0.75kgm²
α = 37.5rad/s²
r = 3.30cm = 0.033m
F = 0.75(37.5)/0.033
F = 28.125/0.033
F = 852.27N
The force exerted by the muscle is 852.27N