Answer:
Workdone = 20 Joules
Explanation:
Given the following data;
Force = 10N
Extension, e = 4cm to meters = 4/100 = 0.04 meters
Workdone extension = 40cm to meters = 40/100 = 0.4 meters
To find the work done;
First of all, we would find the spring constant using the formula;
Force = spring constant * extension
10 = spring constant * 0.04
Spring constant = 10/0.04
Spring constant = 250 N/m
Next, we find the work done;
Workdone = ½ke²
Where;
k is the spring constant.
e is the extension.
Substituting into the formula, we have;
Workdone = ½ * 250 * 0.4²
Workdone = 125 * 0.16
Workdone = 20 Joules
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
Potential energy =
(mass) x (gravity) x (height above the reference level) .
Relative to the bottom of the cliff, the potential energy
at the top of the cliff is
(25kg) x (9.8 m/s²) x (30 meters)
= (25 x 9.8 x 30) kg-m²/s²
= 7,350 joules .
Kinetic energy = (1/2) x (mass) x (speed²)
The rock's kinetic energy at the bottom is
the same as its potential energy at the top.
7,350 joules = (1/2) x (25 kg) x (speed²)
Divide each side
by 12.5kg : 7,350 joules/12.5 kg = speed²
7,350 kg-m²/s² / 12.5kg = speed²
(7,350 / 12.5) m²/s² = speed²
588 m²/s² = speed²
Take the square root
of each side:
Speed = √(588 m²/s²)
= 24.248... m/s (rounded)
Answer:
191.316 K or -81.684 °C
Explanation:
From general gas law,
P₁V₁/T₁ = P₂V₂/T₂ ................ Equation 1
Where P₁ = Initial pressure, V₁ = Initial volume, T₁ = Initial temperature, P₂ = Final pressure, V₂ = Final volume, T₂ = Final Temperature.
Make T₂ the subject of the equation.
T₂ = P₂V₂T₁/P₁ V₁ ............... Equation 2
Given: P₁ = 5.00×10⁶ Pa, T₁ = 25.0°C = 298 K, P₂ = 1.07×10⁶.
Let: V₁ = y cm³, V₂ = 3y cm³
Substitute into equation 2,
T₂ = (1.07×10⁶×298×3y)/(5.00×10⁶×y)
T₂ =191.316 K.
Hence the final temperature = 191.316 K or -81.684 °C
Answer:
Angular resolution describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.
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