Answer:
1 angstrom = 0.1nm
5000 angstrom = 5000/1 × 0.1nm
<h3>= 500nm</h3>

5000 angstrom = 5000 × 1 × 10^-10
<h3>= 5 × 10^-7 m</h3>
Hope this helps you
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

for first resonant frequency, we have n = 1
Inserting the values


Hz
Answer:
The speed with which the baseball leaves the hand = 20.58 m/s
Explanation:
The time take to reach highest height during a projectile's flight is given by
t = (u sin θ)/g
u = initial velocity of the baseball = ?
θ = angle of throw above the horizontal
g = acceleration due to gravity = 9.8 m/s²
1.05 = (u sin 30)/9.8
u = (1.05 × 9.8)/0.5
u = 20.58 m/s
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.