Answer:
The speed will be "18km/s". A further explanation is given below.
Explanation:
According to the question, the values are:
Wavelength,



As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒ 
or,
⇒ 
Answer:
Explanation:
mass of refrigerator, m = 110 kg
coefficient of static friction, μs = 0.85
coefficient of kinetic friction, μk = 0.59
(a) the minimum force required to just start the motion in refrigerator
F = μs x mg
F = 0.85 x 110 x 9.8
F = 916.3 N
(b) The force required to move the refrigerator with constant speed
F' = μk x mg
F' = 0.59 x 110 x 9.8
F' = 636.02 N
(c) Let a be the acceleration.
Net force = Applied force - friction force
F net = 950 - 636.02
F net = 313.98 N
a = F net / mass
a = 313.98 / 110
a = 2.85 m/s²
Answer:
d = 100.8 ft
Explanation:
As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet
here we know that

so here we have

now again if the speed is increased to 48 mph then let say the stopping distance is "d"
so we will have

now divide the above two equations


1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration
directed downward, initial vertical position
, and initial vertical velocity
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

substituting, we find

2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of

In order to do that, we use again the same SUVAT equation substituting
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

Which means that the heigth of the packet was

Answer:
The rate at which velocity changes with respect to a change in time is called. acceleration.
Explanation: