Answer:
The frequency of sound heard by the boy is 1181 Hz.
Explanation:
Given that,
Frequency of sound from alarm 
Speed = -8.25 m/s
Negative sign show the boy riding away from the car
Speed of sound = 343
We need to calculate the heard frequency
Using formula of frequency
Where, 
= frequency of source
= speed of observer
= speed of source
= speed of sound
Put the value into the formula
here, source is at rest
Hence, The frequency of sound heard by the boy is 1181 Hz.
Answer:
3.4 x 10⁴ m/s
Explanation:
Consider the circular motion of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of electron = 9.1 x 10⁻³¹ kg
v = radial speed
r = radius of circular path = 2 mm = 0.002 m
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
For the circular motion of electron
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Consider the linear motion of the electron :
v' = linear speed
x = horizontal distance traveled = 9 mm = 0.009 m
t = time taken = 
= 
= 4.5 x 10⁻⁷ sec
using the equation
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
Speed is given as
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s
Answer:
He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.
Explanation:
By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.
Answer:
T = 11.93 N
Explanation:
Newton's second law to the puck in the circular path
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in radial direccion (N)
m : puck mass (kg)
a : radial acceleration of the puck (m/s²)
Data:
m = 0.085 kg
L = 0.72 m = R : radium of the circular path (m)
θ= one revolution = 2Π rad
t= 0.45 s
Angular speed of the puck
ω = θ/t
ω = 1 rev/0.45 s = (2π/0.45) rad/s
ω = 13.96 rad /s
Radial acceleration or centripetal
a = ω²*R
a = (13.96) ²* (0.72)
a = 140.3 m/s²
Magnitude of the tension in the string (T)
We apply the Formula (1)
∑F = m*a
T = (0.085 kg )* (140.3 m/s² )
T = 11.93 N
In atmospheric science, surface pressure<span> is the atmospheric </span>pressure<span> at a location on Earth's </span>surface<span>. It is directly proportional to the mass of air over that location. For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of </span>surface pressure<span>.
The answer is decrease more slowly
</span>
