Answer:
B is the difference قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *
Negative charge (it has more protons than electrons)
Armature is the correct answer.
Answer:
(d) 2840 A
Explanation:
Threshold frequency can be be calculated using expresion below
(λ)= hc/ Φ
(λ) sodium/ (λ)tungsten = Φtungsten/ Φsodium
But (λ) sodium is given as 5460 A,
Φsodium = 2.3 eV
Φtungsten= 4.4 eV
If we substitute the values we can find
(λ)tungsten.
(λ)tungsten= [(λ) sodium × Φsodium]/ Φtungsten
=( 5460× 2.3)/4.4
= 2854A
Hence, threshold wavelength of tungsten is 2854A
Answer:
P = 981 N
Explanation:
Given
Angle of the incline θ = 10°
Angle of the towing force φ =20°
Weight of the crate W = 3433.5 N
The coefficient of static friction µ = 0.5
Solution
Forces Acting along the ramp
Forces acting perpendicular to the ramp
Substituting the value of N we get
![\mu[Wcos10^{0}-Psin30^{o}] =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- \mu Psin30^{o} =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- Wsin10^{0}=Pcos30^{o}+\mu Psin30^{o} \\\\P=W\frac{\mu cos10^{0}- sin10^{0}}{cos30^{o}+\mu sin30^{o}} \\\\P=3433.5\frac{0.5 \times cos10^{0}- sin10^{0}}{cos30^{o}+0.5 \times sin30^{o}}\\\\P=980.66\\\\ P = 981 N](https://tex.z-dn.net/?f=%5Cmu%5BWcos10%5E%7B0%7D-Psin30%5E%7Bo%7D%5D%20%3DPcos30%5E%7Bo%7D%2BWsin10%5E%7B0%7D%5C%5C%5C%5C%5Cmu%20Wcos10%5E%7B0%7D-%20%5Cmu%20Psin30%5E%7Bo%7D%20%3DPcos30%5E%7Bo%7D%2BWsin10%5E%7B0%7D%5C%5C%5C%5C%5Cmu%20Wcos10%5E%7B0%7D-%20Wsin10%5E%7B0%7D%3DPcos30%5E%7Bo%7D%2B%5Cmu%20Psin30%5E%7Bo%7D%20%5C%5C%5C%5CP%3DW%5Cfrac%7B%5Cmu%20cos10%5E%7B0%7D-%20sin10%5E%7B0%7D%7D%7Bcos30%5E%7Bo%7D%2B%5Cmu%20sin30%5E%7Bo%7D%7D%20%5C%5C%5C%5CP%3D3433.5%5Cfrac%7B0.5%20%5Ctimes%20cos10%5E%7B0%7D-%20sin10%5E%7B0%7D%7D%7Bcos30%5E%7Bo%7D%2B0.5%20%5Ctimes%20%20sin30%5E%7Bo%7D%7D%5C%5C%5C%5CP%3D980.66%5C%5C%5C%5C%20P%20%3D%20981%20N)
