Answer:
500 g
Explanation:
Specific heat of water = 1 j/g-c
Heat given up by hot water = 200(75-25)(1) = 10 000 j
this is the heat GAINED by the cold water
10000 j = x ( 25 -5)(1)
x =500g
A snowboarder on a slope starts from rest and reaches a speed of 4.2 m/s after 7.3 s
The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².
What is acceleration?The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 0 m/s
Final velocity, v= 4.2 m/s
Time elapsed, t = 7.3 seconds.
To find ;
Acceleration, a
The acceleration when the change in velocity is observed by the formula as:
a= (v-u)/(t)
Substitute the given values:
a= (4.2-0)/(7.3)
a=(4.2)/(7.3)
a= 0.575 m/s²
Hence, the car's acceleration will be 0.575 m/s².
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Places rich in limestone and gypsum have hard water . why ??
Answer:
Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.
Explanation:
Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.
Answer:
Hard water indicates a high amount of minerals.
Explanation:
This most often occurs when natural water sources pass through mineral deposits such as limestone or chalk. The process leads to a relatively high level of calcium, iron, and magnesium.
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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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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The pressure of a gas (relative to vacuum) inside a rigid container at 0 °C is:
A) The same as at 25 °C.
B) Not zero, but somewhat less than it would be at 25 °C.
C) zero
The pressure of a gas inside a rigid container at 0 °C will not be zero, but will somewhat be less than it would be at 25 °C.
Temperature and pressure of gasesAccording to Charle's law of gases, the pressure of a gas is directly proportional to its temperature. This is based on the condition that the volume of the gas does not change.
Thus, for a gas inside a rigid container, the pressure at 0 °C may not be zero. However, the pressure will be less than it would be if the temperature were to be 25 °C.
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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]
The value of acceleration due to gravity (g) on a point 10,000 kilometers above sea level is about 1.49 meters/second2. How much will an object, which weighs 98 newtons on the surface of Earth, weigh on this point? The value of acceleration due to gravity on Earth is 9.8 meters/second2.
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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