Answer:
n (a neutron)
Explanation:
For a chemical element:
- The lower subscript indicates the atomic number (the number of protons)
- The upper subscript indicates the mass number (the sum of protons and neutrons in the nucleus)
In the reaction described in the problem, we see that a gamma photon hits a nucleus of Calcium-40, which has
Z = 20 (20 protons)
A = 40 (40 protons+neutrons)
Which means that the number of neutrons is n = A - Z = 40 - 20 = 20
After the reaction, we have a nucleus of Calcium-39, which has
Z = 20 (20 protons)
A = 39 (39 protons+neutrons)
Which means that the number of neutrons is n = A - Z = 40 - 39 = 19
So, the nucleus has lost 1 neutron, which is the particle missing in the reaction.
Answer:
Term 1 = (0.616 × 10⁻⁵)
Term 2 = (7.24 × 10⁻⁵)
Term 3 = (174 × 10⁻⁵)
Term 4 = (317 × 10⁻⁵)
(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.
Explanation:
(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²
mean measurements
Voltage, V = (403 ± 1) V,
σᵥ = 1 V, V = 403 V
Current, I = (2.35 ± 0.01) A
σᵢ = 0.01 A, I = 2.35 A
Coils radius, R = (14.4 ± 0.3) cm
σʀ = 0.3 cm, R = 14.4 cm
Curvature of the electron trajectory, r = (7.1 ± 0.2) cm.
σᵣ = 0.2 cm, r = 7.1 cm
Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)
Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)
Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)
Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)
The relative value of the e/m ratio is a sum of all the calculated terms.
(σ ₑ/ₘ) / (e/m)
= (0.616 + 7.24 + 174 + 317) × 10⁻⁵
= (498.856 × 10⁻⁵)
= (499 × 10⁻⁵) to the appropriate significant figures.
Hope this Helps!!!
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
Answer:
1.6 secs
Explanation:
In a circus act, an acrobat upwards from the surface of a trampoline
At that same moment another acrobat perched 9.0m above him
A ball is released from rest
While still in motion the acrobat catches the ball
He left the ball with a trampoline of 5.6m/s
Since the ball is falling downwards from a distance then acceleration will be negative
a= -9.8
s= d
s= 1/2at^2
= 1/2 × (-9.8)t^2
= 0.5× (-9.8)t^2
d = -4.9t^2
Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows
9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2
9 - 4.9t^2= 5.6t + (-4.9)t^2
9 - 4.9t^2= 5.6t - 4.9t^2
9= 5.6t
t= 9/5.6
t= 1.6 secs
We are given information:
If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2
Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s
Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s
Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s
