The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
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Answer:
H = 0.673
Explanation:
given,
side of cubical crate = 0.74
weight of the crate = 600 N
magnitude of force = 330 N
the Horizontal distance of its Center of mass
= 0.74/2
= 0.37
Let the required Height be H
By Balancing the Torques, we get
H x 330 N = 0.37 x 600
330 H = 222
H = 0.673
hence, the height above the floor where force is acting is equal to 0.673 m
Answer:
Explanation:
Given

mass of core
Average specific heat 
And rate of increase of temperature =
Now
P=

Thus ![\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Btex%5D%5Cfrac%7B1.60%5Ctimes%2010%5E5%5Ctimes%200.3349%7D%7B150%5Ctimes%2010%5E6%7D)

The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
213 s
Explanation:
Slope is the ratio of change in vertical distance to change in horizontal distance.
Slope = vertical height / horizontal height
Therefore:
6.4% = vertical height / 12.42
vertical height = 6.4% * 12.42
vertical height = 0.8 miles
The distance travelled by the car (s) is:
s² = 0.8² + 12.42²
s² = 154.9
s = 12.45 miles
Acceleration (a) = 2.93 ft/s^2 = 0.00055 mile/s²
initial velocity (u) = 0, final velocity = 203 mph
Using:
s = ut + 0.5at²
12.45 = 0.5(0.00055)t²
t =213 s