Answer:
5.831 m/s
Explanation:
According to the work-energy law,
Work done between two points = Change in kinetic energy between the two points.
Since the plastic ball is initially at rest, its initial kinetic energy is 0 since the initial velocity = 0
Work done by the spring = ∫ F.dx
The spring is compressed by 10 cm, so, we integrate from -0.1 m to 0 m
Fₓ(x) = (-30.0 N/m)x+ (60.0 N/m²)x²
F = -30x + 60x²
W = ∫ F.dx = ∫ (-30x + 60x²) dx
W = [- 15x² + 20x³]⁰₋₀.₁ = 0 - [- 15(0.01) + 20(-0.001)] = 0.17 J
W = ΔKE
ΔKE = (mv²/2) - 0
mv²/2 = 0.17
m = 10 g = 0.01 kg
0.01 v² = 0.34
v² = 34
v = 5.831 m/s
Answer:
Lifetime = 4.928 x 10^-32 s
Explanation:
(1 / v2 – 1 / c2) x2 = T2
T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225
T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6
T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s
To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.
To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.
Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.
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Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.
Answer:
B. Energy increases with decreasing wavelength and increasing frequency.
Explanation:
The internal resistance of the battery is 0.5 ohms.
To calculate the internal resistance of the battery, we use the formula below
Formula:
Where:
Make r the subject of the equation
From the question,
Given:
Substitute these values into equation 2
Hence, The internal resistance of the battery is 0.5 ohms.
Learn more about internal resistance here: brainly.com/question/14883923
Answer:
The internal resistance is
Explanation:
From the question we are told that the resistance of
The resistance of the resistor is
The voltage is
The magnitude of the voltage fall is
Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as
substituting values
The internal resistance of the battery is mathematically represented as
substituting values
Answer:
The correct answer is "21195 N".
Explanation:
The given values are:
Tensile strength,
= 3000 MN/m²
Diameter,
= 3.0 mm
i.e.,
= 3×10⁻³ m
Now,
The maximum load will be:
=
On substituting the values, we get
=
=
=
The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.
The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.
We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.
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Answer:
the statements, the correct one is A
a downward force of gravity and an upward force exerted by the surface
Explanation:
When the disc is hit, a thrust force is exerted in the direction of movement, at the moment the disc moves this force loses contact and becomes zero.
When the movement is already established there are two main forces: gravity that acts downwards and the reaction force to the support of the disk called normal that acts upwards.
As it is not mentioned that there is friction, this force that opposes the movement is zero.
Analyzing the statements, the correct one is A