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:
it's all around you and it can't be destroyed
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
