5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting = 
Height of lifting = 
= 5m
The vertical height through which the object was lifted is 5m
learn more:
Work done brainly.com/question/9100769
#learnwithBrainly
Given the following in the problem:
Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s
Get the wavelength of the sound by using the formula:
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2
Get the path length difference to the point from the two speakers
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47
x = delta / lambda
If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.
Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference.
Answer:
d. Not enough information is given to answer this question.
Explanation:
From first law of thermodynamics
Q= W + ΔU
Q=Heat ,W= Work , ΔU=Change in internal energy
If work done by the gas :
It means that W and Q both are positive
Q- W = ΔU
Ii Q > W ,then temperature of the gas will increase.
If Q< W ,Then temperature of the gas will decreases.
If work done on the gas:
Q positive but W will be negative
Q- W = ΔU
Q= W or Q>W or Q< W ,then temperature of the gas will increase.
There are three cases because they did not give any information about the work.That is why option d is correct.
Do you have any answer choices?
Answer:
10.6 mA
Explanation:
t = time interval = 1.00 s
q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C
n₁ = number of Na⁺ ions = 2.68 x 10¹⁶
q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C
n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶
q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C
i₁ = Current due to Na⁺ ions = 
= 
= 0.004288 A
i₂ = Current due to Cl⁻ ions = 
= 
= 0.006272 A
Current passing between the electrodes is given as
i = i₁ + i₂
i = 0.004288 + 0.006272
i = 0.01056 A
i = 10.6 x 10⁻³ A
i = 10.6 mA
