Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + 
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
Answer:
Correct answer: Ek = 2,028.6 J
Explanation:
Ek = m v²/2 = 0.023 · 220² = 0.023 · 176,400/2 = 2,028.6 J
God is with you!!!
Answer:
Explanation:
energy emitted by source per second = .5 J
Eg = 1.43 eV .
Energy converted into radiation = .5 x .12 = .06 J
energy of one photon = 1.43 eV
= 1.43 x 1.6 x 10⁻¹⁹ J
= 2.288 x 10⁻¹⁹ J .
no of photons generated = .06 / 2.288 x 10⁻¹⁹
= 2.6223 x 10¹⁷
wavelength of photon λ = 1275 / 1.43 nm
= 891.6 nm .
momentum of photon = h / λ ; h is plank's constant
= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹
= .0074 x 10⁻²⁵ J.s
Total momentum of all the photons generated
= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷
= .0194 x 10⁻⁸ Js
b ) spectral width in terms of wavelength = 30 nm
frequency width = ?
n = c / λ , n is frequency , c is velocity of light and λ is wavelength
differentiating both sides
dn = c x dλ / λ²
given dλ = 30 nm
λ = 891.6 nm
dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²
= 11.3 x 10¹² Hz .
c )
10 nW = 10 x 10⁻⁹ W
= 10⁻⁸ W .
energy of 50 dB
50 dB = 5 B
I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s
I = I₀ x 10⁵
= 10⁻¹² x 10⁵
= 10⁻⁷ W
= 10 x 10⁻⁸ W
power required
= 10⁻⁸ + 10 x 10⁻⁸ W
= 11 x 10⁻⁸ W.
Answer:
0.006075Joules
Explanation:
The final kinetic energy of the system is expressed as;
KE = 1/2(m1+m2)v²
m1 and m2 are the masses of the two bodies
v is the final velocity of the bodies after collision
get the final velocity using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
0.12(0.45) + 0/12(0) = (0.12+0.12)v
0.054 = 0.24v
v = 0.054/0.24
v = 0.225m/s
Get the final kinetic energy;
KE = 1/2(m1+m2)v
KE = 1/2(0.12+0.12)(0.225)²
KE = 1/2(0.24)(0.050625)
KE = 0.12*0.050625
KE = 0.006075Joules
Hence the final kinetic energy of the system is 0.006075Joules
