Answer:
45 m / s North is a valid vector reading for an object.
Explanation:
Then velocity will be defined by x km / hr North. And, magnitude of velocity defines the speed of the body. Although this tells the speed, but there is no description for the direction, so it's not a vector reading
Answer:
a) (95.4 i^ + 282.6 j^) N
, b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º
Explanation:
a) Let's use trigonometry to break down Jennifer's strength
sin θ = Fjy / Fj
cos θ = Fjx / Fj
Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be
T = 90 -32 = 58º
Fjy = Fj sin 58
Fjx = FJ cos 58
Fjx = 180 cos 58 = 95.4 N
Fjy = 180 sin 58 = 152.6 N
Andrea's force is
Fa = 130.0 j ^
We perform the summary of force on each axis
X axis
Fx = Fjx
Fx = 95.4 N
Axis y
Fy = Fjy + Fa
Fy = 152.6 + 130
Fy = 282.6 N
F = (95.4 i ^ + 282.6 j ^) N
b) Let's use the Pythagorean theorem and trigonometry
F² = Fx² + Fy²
F = √ (95.4² + 282.6²)
F = √ (88963)
F = 298.27 N
tan θ = Fy / Fx
θ = tan-1 (282.6 / 95.4)
θ = tan-1 (2,962)
θ = 71.3º
c) To avoid the movement they must apply a force of equal magnitude, but opposite direction
F' = 298.27 N
θ' = 180 + 71.3
θ = 251.4º
Answer:
The angle is ![\theta = 36.24 ^o](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%2036.24%20%5Eo)
Explanation:
From the question we are told that
The mass is ![m = 0.6 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%200.6%20%5C%20kg)
The radius is ![r = 1.1 \ m](https://tex.z-dn.net/?f=r%20%3D%20%201.1%20%5C%20m)
The speed is ![v = 3.57 \ m /s](https://tex.z-dn.net/?f=v%20%3D%20%203.57%20%5C%20m%20%2Fs)
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
![m * g * h = \frac{1}{2} * m * v^2](https://tex.z-dn.net/?f=m%20%2A%20g%20%20%2A%20%20h%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%2A%20v%5E2)
=> ![h = \frac{1}{2 g } * v^2](https://tex.z-dn.net/?f=h%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%20g%20%7D%20%2A%20%20v%5E2)
Here h is the vertical distance traveled by the mass which is also mathematically represented as
![h = r * sin (\theta )](https://tex.z-dn.net/?f=h%20%20%3D%20%20r%20%2A%20sin%20%28%5Ctheta%20%29)
So
![\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%20sin%20%5E%7B-1%7D%20%5B%20%5Cfrac%7B1%7D%7B2%2A%20g%2A%20r%20%7D%20%2A%20%20v%5E2%5D)
substituting values
![\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%20sin%20%5E%7B-1%7D%20%5B%20%5Cfrac%7B1%7D%7B2%2A%209.8%2A%201.1%20%7D%20%2A%20%20%283.57%29%5E2%5D)
![\theta = 36.24 ^o](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%2036.24%20%5Eo)
Answer:
2.73 K
Explanation:
The Cosmic Microwave Background (CMB) radiation are a type of electromagnetic radiations that are released after the explosion of the big bang. It propagates in all direction in space and is moving further with the expansion of the universe. They are not visible but they are helpful in estimating the age of the universe.
These radiations have a wavelength of about 970 μm (approximately 1 mm).
The temperature of the corresponding blackbody of this radiation is approximately <u>2.73 K(-270.42°C)</u> and its extremely cold.
From definition 1 kWh = 3600000 J
The formula for KE = 1/2 * m * v^2
3600000 = 1/2 * 1000 * v^2
3600000 = 500v^2
v^2 = 7200
v = 84.85 m/s