Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
(1)
where
a is the deleceration
S is the distance covered after a time t
is the velocity at time t
is the initial speed of the car
The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and
(the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve the equation (1) to find a:
And the negative sign means it is a deceleration.
Answer:
The nodes and anti nodes would reverse roles.
Explanation:
I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).
Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....
Answer:
The radiation wavelength is 1.08 X 10⁻¹² m
Explanation:
Frequency is the ratio of speed of photon to its wavelength
F = c/λ
where;
c is the speed of the photon = 3 x 10⁸ m/s
λ is the wavelength of gamma ray = ?
F is the frequency of the gamma ray = 1/T
T is the period of radiation = 3.6x10⁻²¹ s
λ = T*C
λ = 3.6x10⁻²¹ * 3 x 10⁸
λ = 1.08 X 10⁻¹² m
Therefore, the radiation wavelength is 1.08 X 10⁻¹² m
Well it seems that you did not give answer choices, but that its fine since we can use newtons law of universal gravitational, Fg = GM1M2/r^2. So G is the gravitational constant, which is 6.67*10^-11, we can plug in 6*1024 for M1, and 7*1022 for M2, and 3.8*108 for r. Which then we get 1.74 * 10^8 N as the force of attraction between the Earth and the moon.
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries
The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.
At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J
The change in PE is equal to the work done.
Answer: 82.3 J
