Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;

Here, M is the magnification of an astronomical telescope,
is the focal length of the eyepiece lens and
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put
and
in the above expression.

M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.
Answer:
v =2.02
Explanation:
v^2=0.05-4.9
v^2=-4.85
square root both side
v=2.02
^^^^this is a not a perfect square
The answer is C. I hope I helped!
Answer:
Normal force = 8.75 N
Explanation:
given,
frictional force between the steel spatula and the Teflon frying pan=0.350 N
coefficient of friction between material =0.04
normal force = ?
using formula,
Frictional force = coefficient of friction × normal force


Normal force = 8.75 N