Answer: The bug will remain motionless
Explanation:
According to Newton's first Law of Motion (sometimes called Law of Inertia):
<em>An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion.
</em>
In other words:
An object or body will keep its state of motion until an external force changes its state
This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.
In addition, according to his law, an object in rest can be in equilibrium (net force equals to zero), and a moving object can also be in equilibrium, as long as it keeps a constant velocity.
<h2>
This is why the bug, which is at rest will remain at rest, although the ants are simultaneously pulling it in different directions, since the resultant of all these forces is zero.</h2>
Answer:
λ = 1.8 x 10⁻⁷ m = 180 nm
Explanation:
First we find the work function of tungsten by using the following formula:
∅ = hc/λmax
where,
∅ = work function = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λmax = maximum wavelength for photoelectric emission = 230 nm
λmax = 2.3 x 10⁻⁷ m
Therefore,
∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)
∅ = 8.64 x 10⁻¹⁹ J
Now we convert Kinetic Energy of electron into Joules:
K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)
K.E = 2.4 x 10⁻¹⁹ J
Now, we use Einstein's Photoelectric Equation:
Energy of Photon = ∅ + K.E
Therefore,
Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J
Energy of Photon = 11.04 x 10⁻¹⁹ J
but,
Energy of Photon = hc/λ
where,
λ = wavelength of light = ?
Therefore,
11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)
<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>
Answer:
a = 2m/s^2
Explanation:
Force (F) = 100 N
Mass (m) = 50 kg
Here,
F = m×a
100 = 50 × a
a = 100÷50
a = 2m/s^2
Thus, the acceleration on the cart is a = 2m/s^2
-TheUnknownScientist
Answer:
h'=0.25m/s
Explanation:
In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).
So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 
. As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:
notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.
If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:
When solving for r, we get:
so we can substitute this into our volume of a cone formula:
which simplifies to:
So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:
Which simplifies to:
So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)
So we get:
Now we can substitute the provided values into our equation. So we get:
so:
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
