Answer:
A. the wave speed v and Wavelength
Explanation:
Given that
Mass density per unit length=μ
Frequency = f
The velocity V given as
T=Tension
V=Velocity
V= f λ
λ=Wavelength
Therefore to find the tension ,only wavelength and speed is required.
The answer is A.
First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
Answer:
1.2 × 10^27 neutrons
Explanation:
If one neutron = 1.67 × 10^-27 kg
then in 2kg...the number of neutrons
; 2 ÷ 1.67 × 10^-27
There are.... 1.2 × 10^27 neutrons
Answer:
292796435 seconds ≈ 300 million seconds
Explanation:
First of all, the speed of the car is 121km/h = 33.6111 m/s
The radius of the planet is given to be 7380 km = 7380000 m
From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec
If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have
w(vehicle) = 9.78 x w(planet)
w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec
To find the period of the planet's rotation; we use the equation
w(planet) = 2π÷T
Where w(planet) is the angular velocity of the planet and T is the period
From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds
Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds
Answer:
Final speed of car = 12 m/s
Explanation:
We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.
a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s
v = ?
u = 0 m/s
a = 4.0 m/s²
t = 5 s
v = u + at = 0 + 4 x 5 = 20 m/s
b) Then maintains that velocity for 10 s
v = ?
u = 20 m/s
a = 0 m/s²
t = 10 s
v = u + at = 20 + 0 x 10 = 20 m/s
c) Then decelerates at the rate of 2.0 m/s² for 4.0 s
v = ?
u = 20 m/s
a = -2.0 m/s²
t = 4 s
v = u + at = 20 + -2 x 4 = 12 m/s
Final speed of car = 12 m/s
