Answer:
1020.4kg/m³
35m³
Explanation:
Given parameters:
Depth = 0.5m
Pressure exerted = 5000N/m²
Unknown = density of the liquid = ?
Solution:
To find the density of this unknown liquid, we use the expression below:
Pressure of liquid = density x height x acceleration due to gravity
5000 = density x 0.5 x 9.8
Density = 1020.4kg/m³
B.
Density of water = 1000kg/m³
Mass of water = 35000kg
Unknown:
Volume of water = ?
Solution:
The volume of water can be derived from the expression below:
Volume = [tex]\frac{mass}{density}[/tex]
Volume = [tex]\frac{35000}{1000}[/tex] = 35m³
which of the following are not units used to measure energy?
a. joules
b. newtons
c. BTU
d. calories
Answer:
The BTU, or British thermal unit, is actually a measure of heat.
A toy car, mass of 0.025 kg, is traveling on a horizontal track with a velocity of 5 m/s. If
the track then starts to climb upwards, how high up the track can the car reach?
Answer:
1.25 m
Explanation:
This is the vertical height not the distance along the slope.
[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]
The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.
What is kinetic and potential energy?Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.
P.E. = mass × g × height
K.E. = 0.5 × mass × (velocity)²
Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.
Kinetic Energy = Potential Energy
0.5 × mass × (velocity)² = mass × g × height
Cancel mass from both the sides,
0.5 × (velocity)² = g × height
0.5 × (5 m/s)² = 9.81 m/sec² × height
height = 1.2742 meters
Hence, the car will travel 1.2742 meters up.
Learn more about Kinetic and Potential Energy here:
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If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0o. (b) What is the minimum coefficient of friction needed for a frightened dri
Answer:
a) The ideal speed = 16.21 m/s
b) Minimum co-efficient of friction = 0.216
Explanation:
From the given information:
The ideal speed can be determined by considering the centrifugal force component and the gravity component.
[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]
[tex]v = \sqrt {gr \ tan \theta}[/tex]
[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]
= 16.21 m/s
(b)
Let assume that it requires 25 km/h to take the same curve.
Then, using the equilibrium conditions;
[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]
[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]
[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]
[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]
[tex]\mathbf{\mu = 0.216}[/tex]