The height of the roof is <u>3.57m</u>
Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.
Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.
The length of the window s is given by,
The first drop is at the bottom and it takes 5t seconds to reach down.
The height of the roof h is the distance traveled by the first drop and is given by,
the height of the roof is 3.57 m
Answer:
Vx = 35 x cos(13deg)
Vy = 35 x sin(13deg) - gt
(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)
Explanation:
The tiger leaps up, then x and y component of its velocity are:
Vx = Vo x cos(alpha)
Vy = Vo x sin(alpha) - gt
(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)
Hope this helps!
Answer:
You increase the acceleration of the car.
Answer:
B. Containing charged regions
Explanation:
The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions
So in the given case, the option b would be contributed to the molecules that have intermolecular forces
hence, the option b is correct
Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
(the focal length is negative for a diverging lens)
is the distance of the object from the lens
Solvign the equation for q, we find
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens
