Answer:
The Earth Revolves around the sun because the Earth Tilts<u>,</u> we have night and day.
Explanation:
Hope this helps! <3
Answer:
Explanation:
According to Newton's third law of motion, forces always act in equal but opposite pairs. Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted. 1.True 2.falues 3.true 4. not really sure on this one
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.
<h3>What is the cosine rule?</h3>
From the cosine rule we know that;
c^2 = a^2 + b^2 - 2abcosC
Since;
a = 0.47 m
b = 0.62 m
c = 0.78 m
Then;
(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC
0.61 = 0.22 + 0.38 - 0.58 cosC
0.61 - ( 0.22 + 0.38) = - 0.58 cosC
0.01 = - 0.58 cosC
C = cos-1(0.01/-0.58)
C = 91 degrees
Using the sine rule;
b/Sin B = c/Sin C
0.62/sinB = 0.78/sin 91
0.62/Sin B = 0.78
B = sin-1 (0.62//0.78)
B = 53 degrees
Angle A is obtained from the sum of angles in a triangle;
180 - (91 + 53)
A = 36 degrees
Learn more about triangle:brainly.com/question/2773823
#SPJ1
Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of
, and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get

Since we have 


Replacing in [2]


A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it