An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given, the Earth has the acceleration due to gravity, g = 9.81 m/s².
Force of gravity W = mass x acceleration due to gravity
98N = m x 9.8m/s²
m = 10 kg
The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².
The acceleration due to gravity and mass is related as
g = GM/R²
where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.
Substituting the values, we get
1.49 m/s² = 6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²
M = 2.23 x 10²⁴ kg
Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg
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The resistivity of pure water is fairly high, 1.8 × 105 Ω·m, whereas the resistivity of sea water is a million times lower, 2 × 10-1 Ω·m. Why does the high salt concentration make sea water significantly less resistive (i.e. more conductive) than pure water?
The high salt concentration make sea water significantly less resistive than pure water due to presence of charged ions in the sea water.
What is resistivity?The resistivity of a substance is the opposition to the flow of charges offered by the substance.
The greater the resistivity of a substance, the lesser its conductivity.
A high salt concentration make sea water highly conductive due to presence of charged ions in the water. The greater conductivity reduces the resistivity of the sea water.
A pure water has no charged ions in the water, thereby decreasing its conductivity and increasing its resistivity.
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what is the energy of a photon with a frequency of 8.34 x 10^14 hz, in joules?
Answer:
E = 5.52 x 10⁻¹⁹ J
Explanation:
The formula for energy of a photon, E, is:
[tex]E = hf[/tex]
where h is Plank's constant = 6.62 x 10 ⁻³⁴ .
Using the formula:
E = 6.62 x 10⁻³⁴ x 8.34 x 10¹⁴
= 5.52 x 10⁻¹⁹ J
The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s.
Find
A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.
(b) The velocity and acceleration at any time t is v = (4ti + j) m/s and a = a = 4i m/s²
(c) The average acceleration in the time interval given in part (a) is 3.98 m/s².
Position of the particlex = at²i + btj
x = 2t²i + tj
Average velocity, at t₁=2sec and t₂=3secΔv = Δx/Δt
x(2) = 2(2)²i + 2j
x(2) = 8i + 2j
|x(2)| = √(8² + 2²) = 8.246
x(3) = 2(3)²i + 3j
x(3) = 18i + 3j
|x(3)| = √(18² + 3²) = 18.248
Δv = (18.248 - 8.246)/(3 - 2)
Δv = 10 m/s
Velocity and acceleration at any time, tv = dx/dt
v = (4ti + j) m/s
a = dv/dt
a = 4i m/s²
Average accelerationv(2) = 4(2)i + j
v(2) = 8i + j
|v(2)| = 8.06 m/s
v(3) = 4(3)i + j
v(3) = 12i + j
|v(3)| = 12.04 m/s
a = (12.04 - 8.06)/(3 - 2)
a = 3.98 m/s²
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A 3,204 kg tree positioned on the edge of a cliff 247 m above the ground breaks away and falls into the valley below which is considered zero potential energy. If the tree’s mechanical energy is conserved, what is the speed of the tree just before it hits the ground in meters/sec?
Let's see
PE is turned to KE as per law of conservation of energy[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]
[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]
[tex]\\ \rm\Rrightarrow v²=4940[/tex]
[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]
