288.51 N is the magnitude of the force that the beam exerts on the hi.nge.
Given
Mass 0f beam = 40 Kg
The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N
Angle between the beam and cable is = 90°
Angle between beam and the horizontal component = 31°
As the system of the beam, hi_nge and cable are in equilibrium.
The magnitude of the force that the beam exerts on the hi_nge can be calculated by -
F =The horizontal component of force + the vertical component of force
F = 86.62 N + 40 × 9.8 × sin 31°
F =86.62 N + 201.89 N
F = 288.51 N
Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.
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1 . W=mass times acceleration due to gravity
60kg times 9.8m/s2
= 588N
2. W=mg
1176N=m times 9.8
m=120kg
3. 1 hour=3600s
24 hours=?
24 times 3600
= 86400 seconds
4. 1000g=1kg
25000g=?
25000 times 1 divide by 1000
=25kg
5. 1000000mg=1kg
123000000=?
123000000 times 1 divide by 1000000
=123 kg
Answer:
7.468 kN
Explanation:
Here the force of 7468 Newton is given.
Some of the prefixes of the SI units are
kilo = 10³
Mega = 10⁶
Giga = 10⁹
The number is 7468.0
Here, the only solution where the number of significant figures is kilo
1 kilonewton = 1000 Newton
So 7468 N = 7.468 kN
Answer:
q2 = -1.61*10^-5 C.
Explanation:
It was given that,
F = 0.985N
q1 = +8.40 X10-6 C
q2 = ?
r = 1.11 m
k = 9 x 10^9 (standard)
It generally follows that, if force is attractive, charge will be negative.
force, F = kq1q2/r^2
0.985 = 9*10^9*8.40*10^-6*q2/1.11^2
75600q2 = 0.985*1.11^2
75600q2 = 1.2136
q2 = 1.2136/75600 = 1.60529
q2 = -1.61*10^-5 C.
