Part a can be solve using the equation of trajectory:
Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]
Where y is the height
X is the length
G is the acceleration due to gravity
Vo Is the initail velocity
a is the angle of trajectory
1.2 = 1.35 tan(0) +
(9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]
Solve for V0 = 2.729 m/s
b. can be solve using the formula
v = sqrt(2gy)
= sqrt ( 2*1.2*9.81)
= 4.852 m/s going
down ( 0 degree from the horizontal)
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Answer:
<em>The mass of the object is 745000 units of the sun</em>
Explanation:
We know that the centripetal force with which the stars orbit the object is represented as
= 
and this centripetal force is also proportional to
= 
where
m is the mass of the stars
M is the mass of the object
v is the velocity of the stars = 10^6 m/s
r is the distance between the stars and the object = 10^14 m
k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
We can equate the two centripetal force equations to give
=
which reduces to
= 
and then finally
M = 
substituting values, we have
M = 
= 1.49 x 10^36 kg
If the mass of the sun is 2 x 10^30 kg
then, the mass of the the object in units of the mass of the sun is
==> (1.49 x 10^36)/(2 x 10^30) = <em>745000 units of sun</em>
Answer:
The correct answer is - the sound waves make vibration that travels through the string.
Explanation:
When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.
The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.
You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them
Answer:
transition metal, and inner transition metals groups are numbered 1-18 from left to right
Answer:
7/150
Explanation:
The following data were obtained from the question:
Object distance (u) = 75cm
Image distance (v) = 3.5cm
Magnification (M) =..?
Magnification is simply defined as:
Magnification (M) = Image distance (v)/ object distance (u)
M = v /u
With the above formula, we can obtain the magnification of the image as follow:
M = v/u
M = 3.5/75
M = 7/150
Therefore, the magnification of the image is 7/150.
