Answer:
T = 540 N (to two significant digits)
Explanation:
Let the crate dimension L be from strap attachment to floor contact
Let T be the strap tension
sum moments about the floor contact point to zero
mg[½Lcos25] - Tsin61[Lcos25] + Tcos61[Lsin25] = 0
L is common to all terms, so divides out.
½(71)(9.8)cos25 = T(sin61cos25 - cos61sin25)
T = (71)(9.8)cos25 / (2(sin61cos25 - cos61sin25))
T = 536.428020...
Answer:
2.5 x 10^{5} J
Explanation:
weight = 5,000 N
coefficient of friction = 0.05
distance = 1000 m
how much work is done by the dogs pulling the sledge
work done = force x coefficient of friction x distance
work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J
We know that velocity is equal to the total displacement of an object over time.

Deriving from that equation, we can say that:

Okay, so here it goes:

The bicycle took 25.02 seconds to displace at 58.3 meters.
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

Here,
D is diameter of the eye


The angle that relates the distance between the lights and the distance to the lamp is given by,

For small angle, 
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle 
Therefore,



Therefore the distance is 5.367km.