Two blocks A and B with mA = 2.9 kg and mB = 0.87 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 6 N. Determine the tension in the string connecting the two blocks.
Answer:
Tension = 1.38 N
Explanation:
Given that:
Mass of block A [tex]m_A[/tex] = 2.9 kg
mass of block B [tex]m_B[/tex] = 0.87 kg
Force F = 6 N
Assume a = acceleration of the blocks.
Then:
[tex]m_A[/tex] (a) + [tex]m_B[/tex] (a) = 6
2.9a + 0.87a = 6
3.77a = 6
a = 6/3.77
a = 1.59 m/s²
Suppose T to be the tension in the string.
If we take a look at the forces acting on the first block, then:
F - T = [tex]m_A[/tex] (a)
T = F - [tex]m_A[/tex] (a)
T = 6 - 2.9(1.59)
T = 6 - 4.62
T = 1.38 N
A ball is spun around in circular motion such that its frequency is 10 Hz.
a. What is the period of its rotation?
b. How much time will be required to complete 100 rotations?
Answer:
a = 0.1 s b. 10 s
Explanation:
Given that,
The frequency in circular motion, f = 10 Hz
(a) Let T is the period of itsrotation. We know that,
T = 1/f
So,
T = 1/10
= 0.1 s
(b) Frequency is number of rotations per unit time. So,
[tex]t=\dfrac{n}{f}\\\\t=\dfrac{100}{10}\\\\t=10\ s[/tex]
Hence, this is the required solution.
Which element is the biggest contributor to Climate Change?
Answer:
carbon dioxide (CO2)
Explanation:
the burning or combustion of these fossil fuels creates gases that are released into the atmosphere. Of these gases, carbon dioxide (CO2) is the most common and is the gas most responsible for exacerbating the green- house effect that is changing global climate patterns.
A spherical light bulb dissipates 100W and is of 5cm diameter. Assume the emissivity is 0.8 and the irradiation is negligible. What is the surface temperature of this spherical light bulb
Answer:
[tex]T=728.9K[/tex]
Explanation:
Power [tex]P=100W[/tex]
Diameter [tex]d=5[/tex]
Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]
Emissivity [tex]e=0.8[/tex]
Generally the equation for Area of Spherical bulb is mathematically given by
[tex]A=4\pi r^2[/tex]
[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]
[tex]A=7.85*10^{-3}m^2[/tex]
Generally the equation for Emissive Power bulb is mathematically given by
[tex]E=e\mu AT^4[/tex]
Where
[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]
Therefore
[tex]T^4=\frac{E}{e\mu A}[/tex]
[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]
[tex]T=^4\sqrt{2.80*10^{11}}[/tex]
[tex]T=728.9K[/tex]
