Work = Force x distance
(10 pounds)(2 feet)
Work = 20 foot-pounds of work
hope this helps :)
Kinetic energy = momentum^2 / 2 x mass
Mass = (momentum^2/ Kinetic energy) / 2
Mass = (25^2 / 275) / 2
Mass = 1.136 kg.
momentum = mass x velocity
velocity = mass / momentum
velocity = 1.136 / 25
velocity = 0.04544 m/s
Let u = the speed of the car at the instant when braking begins.
The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.
Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
u = 27.2546 m/s
Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
= 2.7696 m/s
Answer: 2.77 m/s (nearest hundredth)
Answer:
0.04455 Hz
Explanation:
Parameters given:
Wavelength, λ = 6.5km = 6500m
Distance travelled by the wave, x = 8830km = 8830000m
Time taken, t = 8.47hours = 8.47 * 3600 = 30492 secs
First, we find the speed of the wave:
Speed, v = distance/time = x/t
v = 8830000/30492 = 289.58 m/s
Frequency, f, is given as velocity divided by wavelength:
f = v/λ
f = 289.58/6500
f = 0.04455 Hz
Answer:
13.309 m/s²
Explanation:
Length from shoulder to hand, l = 30 cm = 0.3 m
initial velocity, u = 1 m/s
final velocity, v = 2.5 m/s
time, t = 3 s
Let the tangential acceleration is a.
by using first equation of motion
v = u + at
2.5 = 1 + 3 a
a = 0.5 m/s²
Let the centripetal acceleration is a'.
a' = v'²/l
a' = 2 x 2 / 0.3
a' = 13.3 m/s²
The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by
A = 13.309 m/s²
