Answer:
5208N
Explanation:
force=mass×acceleration
mass(kg)=1,302
acceleration(m/s²)=4
f= m × a
f= 1302 × 4
f=5208N
Which option gives an object's volume in SI units?O A. 2,6 m3OB. 4.3 kgOC. 3.4 LD. 5.5 K
ANSWER:
A. 2.6 m^3
STEP-BY-STEP EXPLANATION:
The volume of an object is given by the cube of its length.
In SI the unit for length is the meter, therefore, in the case of volume it would be cubic meters.
This means that the correct answer is: A. 2.6 m^3
A bicycle with a mass of 10.0 kg is pushed up an incline of 30 ° with respect to the horizon. Friction is assumed to be negligible. a) Draw the free body diagram. b) It determines the intensity of the force which, applied horizontally, makes the bicycle move forward at a constant speed. c) Determines the intensity of the force which, applied horizontally, moves the bicycle forward with an acceleration of 100 m / s²
a) The free body diagram is drawn in the explanation section
b) The intensity of the force applied horizontally which makes the bicycle move forward at a constant speed = 49.05 N
c) The intensity of the force applied horizontally which makes the bicycle move forward with an acceleration of 100 m/s² = 1049.05N
Explanation:The mass of the bicycle, m = 10.0 kg
Angle of inclination, θ = 30°
The free body diagram of the illustration is drawn below
The bicyce is pushed up an incline of 30°
The weight of the bicycle acts downward (because the weight of every object acts downward)
Since the bicycle is pushed up an inclined of 30°, the force has to be resolved to the vertical(mgcosθ) and horizontal(mgsinθ)
b) The net force on the bicycle is:
[tex]\sum F=mg\sin \theta+ma[/tex]If the bicycle moves at constant speed, the acceleration is 0 m/s²
a = 0m/s²
[tex]\begin{gathered} \sum F=10(9.81)\sin 30+10(0) \\ \sum F=10(9.81)(0.5) \\ \sum F=49.05N \end{gathered}[/tex]c) The intensity of the force applied horizontally which moves the bicycle forward with an acceleration of 100 m/s²
[tex]\begin{gathered} \sum F=10(9.81)\sin 30+10(100) \\ \sum F=49.05+1000 \\ \sum F=1049.05N \end{gathered}[/tex]If frictional forces and air resistance were acting upon the falling ball in #1 would the kinetic energy of theball just prior to striking the ground be more, less, or equal to the value predicted in #1?
Given that the ball is falling in two different situations.
Kinetic energy is the energy that an object possesses due to its motion and it is proportional to the square of the velocity of the object.
The frictional force and the air resistance are the forces that will be opposing the motion of the ball. That is these forces act in the opposite direction of the motion of the ball thus reducing the net acceleration acting on the ball.
As the acceleration decreases the final velocity of the ball decreases.
Thus the kinetic energy of the ball just prior to striking the ground will be less than before.
Calculate the mass of an object with a weight of 467 N.
We will have that the mass of the object is:
[tex]\begin{gathered} m=\frac{467N}{9.8m/s^2}\Rightarrow m=\frac{2335}{49}kg \\ \\ \Rightarrow m\approx47.65kg \end{gathered}[/tex]So, the mass of the object is 2335/49 kg, that is approximately 47.65 kg.
Calculate the current needed to carry the 2 000 MW generated at alarge power station if the distribution voltage is kept at 22 kV, thegenerator output voltage.
Given data
*The given power is P = 2000 MW = 2000 × 10^6 W
*The given distribution voltage is V = 22 kV = 22 × 10^3 V
The formula for the current needed to carry the 2000 MW generated at a large power station is given as
[tex]I=\frac{P}{V}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} I=\frac{2000\times10^6}{22\times10^3} \\ =90.90\times10^3\text{ A} \end{gathered}[/tex]Hence, the current needed to carry the 2000 MW generated at a large power station is I = 90.90 × 10^3 A
when a car driver reached the crest of a hill, she saw the stop light turn red and slammed on the brakes. the car slowed down as it moved to the bottom of the hill. is the KE and PE increasing, decreasing, or staying the same?
As the brakes are opposing the motion of the car and trying to decrease the velocity of the car while moving down the hill.
Thus, the kinetic energy of the car is decreasing as the car is moving down the hill.
The height of the car is also decreasing with the motion of the car down the hill, thus, the potential energy of the car is also decreasing.
The brakes are applying the frictional force on the car which is converting their energy into heat energy. The thermal energy of the car increases as the car moves down the hill.
Thus, the kinetic energy and potential energy of the car decreases while moving down the hill.
If you have ever been to the beach you will find that the sand heats up and cools down faster than the water. What can you predict about the heat capacity of water and sand based on this result.
Water has a higher heat capacity than sand
The change in temperature is faster in sand than in water
In deep water, a tsunami moves very fast and has a long wavelength and a small amplitude (see A in picture). As it enters shallower water, it slows down, and the wavelength decreases. This causes the wave to become much taller (see B in picture).As waves slow down, they start to bunch together, so they have a shorter wavelength than before. This can also be explained by the wave equation v = f x λ (speed = frequency x wavelength), which shows that, when a wave’s speed decreases, it must have a shorter wavelength than before – slowing down won’t change the wave’s frequency.Having a shorter wavelength means that the waves get higher. You could think of the shortened wave as being ‘squashed’ sideways – the water in the wave has to get higher because there’s not as much room for it within the shorter wavelength. It’s a bit like squeezing a toothpaste tube – all the toothpaste is forced upwards.
Given:
The wavelength of the wave decreases as it gets closer to the shore.
The speed of the wave decreases as it gets closer to the shore.
The waves get taller as it gets closer to the shore.
To find:
How does a change in velocity cause a change in amplitude?
Explanation:
The amplitude of a wave is described as the distance of the highest point of vibration or the oscillation of the wave from its mean position.
Given that as the velocity of the speed decreases the waves get taller. That is as the velocity decreases the distance of the highest point of vibration of the wave from its equilibrium or mean position increases.
Final answer:
Thus as the velocity of the wave decreases, its amplitude increases.
Here are three stress and strain graphs shown on approximately the same scale. Match each graph to one of either copper, glass or rubber. Give reason for your choices, perhaps making reference to the grey circular points
For the three graphs A,B and C in the graph C we see that the strain is larger for a small value of stress.
This type of graph is for the polymers (rubber).
The graph A has smallest amount of strain for stress in comparison with B.
Upto the grey circle stress and strain are proportional .
Graph A is for Glass and B for Copper.
A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of sea water is 1.03 x 10^3 kg/m^3. A) calculate the magnitude of the force in newtons pressing on the hatch from outside by the sea water given it is circular and 0.25m in diameter. The air pressure inside the submarine is 1.00atm (101,325 Pa). B) calculate the magnitude of the force, in newtons needed to open the hatch from the inside
Given:
The height (depth) of the hatch from the seawater surface is: h = 25 m
The density of seawater is: ρ = 1.03 × 10³ kg/m³
The diameter of the hatch is: d = 0.25 m
The pressure inside the submarine is: Ps = 1 Atm = 101325 Pa
To find:
The magnitude of the force pressing the hatch from the outside by the seawater and the magnitude of force to open the hatch from the inside.
Explanation:
The net pressure on the hatch area is given as:
[tex]P_{net}=\frac{F_{net}}{A}..........(1)[/tex]Here, Pnet is the net pressure on the hatch of area A, and Fnet is the net force acting on the hatch from the outside.
The net pressure Pnet is given as:
[tex]P_{net}=P_{atm}+P_w-P_s[/tex]Here, Patm is the atmospheric pressure above the seawater surface, Pw is the pressure of the water and Ps is the pressure inside the submarine.
The atmospheric pressure Patm of 1 Atm will contribute to the net pressure on the hatch from the outside. The atmospheric pressure Patm will have no effect as it will cancel out with the pressure Ps of equal magnitude as it is in an opposite direction from inside the submarine.
Substituting the values of the atmospheric pressure Patm and the pressure inside the submarine Ps in the above equation, we get:
[tex]\begin{gathered} P_{net}=1\text{ Atm}+P_w-1\text{ Atm} \\ \\ P_{net}=P_w..........(2) \end{gathered}[/tex]Thus the net pressure on the hatch will be the pressure of the seawater only.
Now, the net pressure Pnet can be calculated as:
[tex]P_{net}=h\rho g[/tex]Here, g is the acceleration due to the gravity having a value of 9.8 m/s².
Substituting the values in the above equation, we get:
[tex]\begin{gathered} P_{net}=25\text{ m}\times1.03\times10^3\text{ kg/m}^3\times9.8\text{ m/s}^2 \\ \\ P_{net}=252350\text{ N/m}^2\text{..........\lparen3\rparen} \end{gathered}[/tex]The area of the hatch can e calculated as:
[tex]\begin{gathered} A=\pi r^2 \\ \\ A=\pi\times(\frac{0.25}{2})^2 \\ \\ A=0.0491\text{ m}^2..........(4) \end{gathered}[/tex]Substituting equation (3) and equation (4) in equation (1) and rearranging it, we get:
[tex]\begin{gathered} 252350\text{ N/m}^2=\frac{F_{net}}{0.0491\text{ m}^2} \\ \\ F=252350\text{ N/m}^2\times0.0491\text{ m}^2 \\ \\ F=12390.39\text{ N} \end{gathered}[/tex]A force of 12390.39 Newtons acts on the hatch by the seawater above it. To open the hatch from the inside, an equal magnitude of force is needed to apply from inside the submarine. Thus, a force of 12390 Newtons is required to open the hatch from the inside.
Final answer:
The magnitude of force pressing the hatch from the outside by the seawater is 12390.39 Newtons.
The magnitude of force required to open the hatch from inside the submarine is 12390.39 Newtons.
(A) The magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N.
(B) The force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.
What is force?In the body, force is the result of either a push or a pull. The three main categories of forces are friction, nuclear, and gravitational. For instance, when a hand strikes a wall, the hand puts force on the wall and the wall also exerts a force on the hand. Newton was given a variety of laws to help him understand force.
Given :
The density of seawater, ρ = 1.03 x 10³ kg / m³ ,
The height of the hatch of the submarine, h= 25 m,
The gravitational acceleration, g = 9.8 m / s² ,
The radius of the hatch, r = 0.25 m
Calculate the area of the hatch by the formula given below,
A = π × r² = π × (0.125 m)² [r = d / 2]
A = 4.9 x 10 ⁻² m²
(A) Calculate the force of the water by the following formula,
[tex]P = F / A[/tex]
F = P × A (P = ρ × g × h)
F = ρ × g × h × A
Substitute the values,
F = 1.03 x 10³ kg / m³ × 9.8 m / s² × 25 m × 4.9 x 10 ⁻² m²
F = 12365.15 N or 1.24 × 10 ⁴ N
(B) The force needed to open the hatch = The Force applied by water on the hatch.
Therefore, the magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N, and the force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.
To know more about Force:
https://brainly.com/question/13191643
#SPJ2
In Denver, children bring their old jack o lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. suppose that the tower is 9.0m high and that the bullseye is a horizontal distance of 3.5m from the launch point.if the pumpkin is thrown horizontally what is the launch speed needed to hit the ground? (b) If the jack o lantern is given an initial horizontal speed of 3.3m/s, what are the direction and magnitude of its velocity (c) 0.75s later and (d) just before it lands.
a)
In order to calculate the required horizontal speed, first let's calculate the falling time, using the free-fall formula:
[tex]d=\frac{gt^2}{2}[/tex]For d = 9 and g = 9.8, we have:
[tex]\begin{gathered} 9=4.9t^2 \\ t^2=\frac{9}{4.9} \\ t=1.355\text{ s} \end{gathered}[/tex]Then, let's use this time in the following formula for horizontal distance, so we can calculate the horizontal speed:
[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ 3.5=\text{speed}\cdot1.355 \\ \text{speed}=\frac{3.5}{1.355}=2.58\text{ m/s} \end{gathered}[/tex]b)
Let's calculate the vertical velocity after 0.75 seconds, using the formula:
[tex]\begin{gathered} V=V_0+a\cdot t \\ V=0+9.8\cdot0.75 \\ V=7.35\text{ m/s} \end{gathered}[/tex]The magnitude can be calculated using the Pythagorean Theorem with the horizontal and vertical velocities:
[tex]\begin{gathered} V^2=7.35^2+3.3^2 \\ V^2=54.0225+10.89 \\ V^2=64.9125 \\ V=8.06\text{ m/s} \end{gathered}[/tex]And the direction is given by the arc tangent of the vertical velocity divided by the horizontal velocity (for this, let's use a negative value of vertical velocity, since it points downwards)
[tex]\theta=\tan ^{-1}(-\frac{7.35}{3.3})=\tan ^{-1}(-2.2272)=-65.82\degree[/tex](d)
To find the vertical velocity, let's use Torricelli's equation:
[tex]\begin{gathered} V^2=V^2_0+2\cdot a\cdot d \\ V^2=0^2+2\cdot9.8\cdot9 \\ V^2=176.4 \\ V=13.28\text{ m/s} \end{gathered}[/tex]Calculating the final speed magnitude and orientation, we have:
[tex]\begin{gathered} V^2=13.28^2+3.3^2 \\ V^2=187.2484 \\ V=13.68\text{ m/s} \\ \\ \theta=\tan ^{-1}(-\frac{13.68}{3.3})=\tan ^{-1}(-4.145)=-76.44\degree \end{gathered}[/tex]I have finished most of this question just need help on making sure it’s correct and the last 2
Given
m: mass
m = 50 kg
KE: kinetic energ
PEg: gravitational potential energy
v: speed
Procedure
Point A
KE = 0
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 100 m
PEg = 50000 J
v = 0 m/s
Point B
h = 60 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 60 m
PEg = 30000 J
KE = 20000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot20000\text{ J}}{50\text{ kg}}} \\ v=\sqrt[]{800\text{ }} \\ v=28.28\text{ m/s} \end{gathered}[/tex]Point C
h = 30 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 30 m
PEg = 15000 J
KE = 35000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot35000\text{ J}}{50\operatorname{kg}}} \\ v=37.41\text{ m/s} \end{gathered}[/tex]Point D
h = 60 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 60 m
PEg = 30000 J
KE = 20000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot20000\text{ J}}{50\text{ kg}}} \\ v=\sqrt[]{800\text{ }} \\ v=28.28\text{ m/s} \end{gathered}[/tex]
Point E
h = 0
PEg = 50 kg * 10 m/s^2 * 0
PEg = 0 J
KE = 50000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot50000\text{ J}}{50\text{ kg}}} \\ v=44.72\text{ m/s} \end{gathered}[/tex]A ball of mass 1.86 kilograms is attached to a cord 1.29 meters long and swung in a vertical circle at a constant speed of 5.27 meters per second. What is the centripetal force acting on the ball? Include units in your answer. What is the tension in the cord when the ball is at the bottom of its path? Include units in your answer. What is the tension in the cord when the ball is at the top of its path? Include units in your answer. All answers must be in 3 significant digits.
Explanation
Step 1
Draw
so
a)centripetal force:
the centripetal force is given by.
[tex]\begin{gathered} F=ma \\ F=m\frac{v^2}{r} \\ \text{where } \\ F_{C\text{ }}\text{ is the centripetal force} \\ m\text{ is the mass } \\ v\text{ is the velocty } \\ r\text{ is the radius} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} F=m\frac{v^2}{r} \\ F=1.86\text{ kg }\frac{(\text{ 5.27 }\frac{m}{s})^2}{1.29\text{ m}} \\ F=40.044\text{ Newtons} \end{gathered}[/tex]so, the centripetal force is 40.0446 Newtons
b) What is the tension in the cord when the ball is at the bottom of its path?
to find the tension in bottom, we need to add the weigth of the ball,so
[tex]\begin{gathered} \text{weigth}=\text{ mass}\cdot accelofgravity \\ w=mg \end{gathered}[/tex]hence, the tension would be
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ \end{gathered}[/tex]replace
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ T_{bottom}=40.044\text{ N+(1.86 kg}\cdot9.81\text{ }\frac{\text{m}}{s^2}) \\ T_{bottom}=40.044\text{ N+18.2466 N} \\ T_{bottom}=58.291\text{ N} \end{gathered}[/tex]c)What is the tension in the cord when the ball is at the top of its path?
to find the tension in the top we need to subtract the weigth, so
[tex]\begin{gathered} T_{\text{top}}=m\frac{v^2}{r}-mg \\ replace \\ T_{\text{top}}=40.044\text{ N-18.2466 N} \\ T_{\text{top}}=21.79\text{ Newtons} \end{gathered}[/tex]I hope this helps you
it would be greatly appreciated if you plug in the numbers according to the format of my worksheet.
a. List variables
[tex]\begin{gathered} m_{1i}=\text{ 0.015 kg} \\ v_{1i}=\text{ 0.225 m/s} \\ m_{2i}=\text{ 0.03 kg} \\ v_{2i}\text{ =-0.18 m/s} \\ m_{1f}=\text{ 0.015 kg} \\ v_{1f}=-0.315\text{ kg} \\ m_{2f}=\text{ 0.03 kg} \\ v_{2f}\text{ = ?} \end{gathered}[/tex]b. plug into the equation
[tex]\begin{gathered} m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1f}v_{1f}+m_{2f}v_{2f} \\ 0.015\times0.225+0.03\text{ }\times(-0.18)=0.015\times(-0.315)+0.03(v_{2f}) \end{gathered}[/tex]c. solve for missing
[tex]\begin{gathered} v_{2f}=\frac{0.015\times0.225+0.03(-0.18)-0.015\times(-0.315)}{0.03} \\ =0.09\text{ m/s} \end{gathered}[/tex]A 1 kg mass has a kinetic energy of 1 Joule when its speed isA. 0.45 m/sB. 1.4 m/sC. 1 m/sD. 4.4 m/s
Kinetic energy = 1/2 x mass x velocity^2
Where:
Kinetic energy = 1 J
mass = 1 kg
Isolate v (speed)
KE = 1/2 m v^2
√(2KE/m) = v
v = √(2KE/m)
Replacing with the values given:
v = √(2*1/1)
v= √2 = 1.4 m/s
Answer: 1.4 m/s (B)
I would appreciate if you can help me solve this problem. I’m having trouble with it
The frequency of a wave is given by:
[tex]f=\frac{v}{\lambda}[/tex]Where v is the speed of the wave and lambda is the wavelenght. In this case the speed is 345 and the wavelenght is 0.85 meters; plugging the values given we have:
[tex]\begin{gathered} f=\frac{345}{0.8} \\ f=431.25 \end{gathered}[/tex]Therefore the frequency of the wave is 431.25 Hz
The maximum possible efficiency of a reversible heat engine is 79.46% when the cold temperature is 224.0°C. What is the hot temperature in degrees Celsius?
We have
A heat engine is 79.46%
TC= 224.0°C
TH=?
we have the next formula
[tex]\eta(\text{\%)}=1-\frac{T_C}{T_H}\times100[/tex]we substitute the values
[tex].7946=1-\frac{224}{T_H}\times100[/tex]then we isolate TH
[tex]\begin{gathered} .07946-1=-\frac{T_C}{T_H}\times1 \\ -0.2054=-\frac{T_C}{T_H}\times1 \\ \frac{T_C}{T_H}\times1=-0.2054 \\ \frac{T_C}{T_H}=\frac{0.2054}{1} \\ \frac{T_H}{T_C}=\frac{1}{0.2054} \\ T_H=4.8685\cdot224 \\ T_H=1090.55\text{ \degree{}C} \end{gathered}[/tex]the hot temperature is 1090.55 degrees Celsius.
When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy. Determine if the statement is correct.
Let's determine if the statement ''When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy'' is correct.
Let's first define kinetic and potential energies.
Potential energy is the energy stored in an object due to its position which depends on the relative position of various parts of the system.
Kinetic energy is the energy possesed by an object in motion.
Therefore, before the arrow is released from the bow, it is at rest and the energy is stored.
Hence, it posssess potential energy.
When the arrow is now released from the bow, it is now in motion. Since it is now in motion, the potential energy is now transformed to kinetic energy.
Therefore, the statement ''When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy'' is correct.
How do i solve this problem? Hint: The cannonball is being launched vertically upwards, therefore, there is no initial horizontal speed. The given initial speed will also be the initial vertical speed.
The initial velocity of the ball is given as 36.0 m/s.
The horizontal component of velocity of ball is given as,
[tex]v_x=v\cos \theta[/tex]The ball is projected vertically, therefore, the angle made by ball is 90 degree.
Plug in the known values,
[tex]\begin{gathered} v_x=(36.0m/s)cos90^{\circ} \\ =(36.0\text{ m/s)(0)} \\ =0\text{ m/s} \end{gathered}[/tex]Therefore, the initial horizontal velocity of ball is 0 m/s.
The vertical component of velocity of ball is given as,
[tex]v_y=v\sin \theta[/tex]Plug in the known values,
[tex]\begin{gathered} v_y=(36.0m/s)\sin 90^{\circ} \\ =(36.0\text{ m/s)(1)} \\ =36.0\text{ m/s} \end{gathered}[/tex]Therefore, the initial vertical velocity of the ball is 36.0 m/s.
A spaceship and an asteroid are moving in the same direction away from Earth with speeds of 0.8 c and 0.45 c, respectively. What is the relative speed between the spaceship and the asteroid?
Answer:
0.35c
Explanation:
The relative speed between the spaceship and the asteroid can be calculated as the difference between their speeds, so it is equal to
Relative speed = 0.8c - 0.45c
relative speed = 0.35c
Therefore, the answer is
0.35c
A laboratory cart with a mass of 7 kilograms moves a distance of 0.5 meters in 3 seconds.Calculate the momentum.
Given that the mass of the cart is m = 7 kg.
The distance is d = 0.5 m.
The time is t = 3 s.
We need to find the momentum, p.
The formula to calculate momentum is
[tex]\begin{gathered} p=mass\times velocity \\ =m\times\frac{d}{t} \end{gathered}[/tex]
Substituting the values, the momentum will be
[tex]\begin{gathered} p=7\times\frac{0.5}{3} \\ =1.167kg\text{ m/s} \end{gathered}[/tex]Thus, the momentum is 1.167 kg m/s.
a block sliding on the ground where uk= 0.193 experiences a 14.7N friction force. what is the mass of the block in kg
Answer:
F = ma
14.7N = m(0.193)
m = 76.3 kg
Which of the following is NOT a requirement for a planet? A. must orbit a star, but is not a star or satellite of another planetB. must be roundC. must clear its orbit of debrisD. must have a moon
Answer:
D. must have a moon
Explanation:
Scientists say that a planet must orbit a star, must be round, and must be big enough so its gravity clears its orbit of other objects. Therefore, the statement that is not a requirement for a planet is
D. must have a moon
For example, Mercury and Venus are planets with no moons.
A rectangular container measuring 34 cm ⨯ 38 cm ⨯ 55 cm is filled with water. What is the mass of this volume of water in kilograms and grams?
We will have the following:
*Kg:
[tex](\frac{1kg}{m^3})(0.34m\ast0.38m\ast0.55m)=0.07106kg/m^3[/tex]*G:
[tex](\frac{1000g}{m^3})(0.34m\ast0.38m\ast0.55m)=71.06g/m^3[/tex]Question 1: Assume that the pendulum of a grandfather clock acts as one of Planck'sresonators. If it carries away an energy of 8.1 x 10-15 eV in a one-quantumchange, what is the frequency of the pendulum? (Note that an energy this smallwould not be measurable. For this reason, we do not notice quantum effects in thelarge-scale world.)
Given:
Energy = 8.1 x 10⁻¹⁵ eV.
Let's find the frequency of the pendulum.
To find the frequency, apply the formula for the energy of a light quantum:
[tex]E=hf[/tex]Where:
E is the energy
h is Planck's constant = 6.63 x 10⁻³⁴ m² kg/s
f is the frequency.
Where:
1 eV = 1.6 x 10⁻¹⁹ J.
Rewrite the formula for f and solve:
[tex]f=\frac{E}{h}[/tex]Thus, we have:
[tex]f=\frac{8.1\times10^{-15}*(1.6\times10^{-19})}{6.63\times10^{-34}}[/tex]Solving further:
[tex]\begin{gathered} f=\frac{8.1\times10^{-15}*(1.6\times10^{-19})}{6.63\times10^{^{-34}}} \\ \\ \\ f=1.95\text{ Hz.} \end{gathered}[/tex]Therefore, the frequency of the pendulum is 1.95 Hz.
ANSWER:
1.95 Hz
What would happen to the pressure and temperature of a box as gas is added to it
When gas is added to the box, the number of molecules in the box will increase.
This will result in an increase in pressure as volume is fixed or constant.
As the box has a fixed volume, this will be an isochoric process.
In an isochoric process, the pressure is directly proportional to temperature, so the temperature will also increase.
Thus, pressure and temperature will increase when gas is added to a box.
If a bullet with mass 2.7 grams is fired from a gun with a speed of 202
m/s into a block of wood. If the kinetic energy is transformed to heat,
what is the increase in temperature of the bullet in degrees C,
assuming the specific heat of the bullet is 234 J/kg/C.
The increase in temperature of the bullet in degrees C is 87.17
Define Specific Heat Capacity?
When a material's temperature rises by 1 K (or 1 °C), or when its mass increases by 1 kg, the specific heat capacity is defined as the amount of heat (J) absorbed per unit mass (kg).
This measurement is expressed as J/(kg K) or J/(kg °C).
The heat that is transferred to the bullet as a result of the conversion of kinetic energy to heat is what causes the bullet's temperature to rise.
Q = cm T,
where Q is the energy transferred, c is the substance's specific heat, m is its mass, and T is the temperature change, describes this relationship.
Given,
mass = 2.7 x 10^-3 kg
speed = 202 m/s
C = 234 J/kg/C.
The kinetic energy of the travelling bullet is,
1/2 m(v^2) = 1/2 x 2.7 x 10^-3 x 202 x 202
= 55.08 J
Energy that raises the temperature = 55.08 J
This is equal to mCΔT
Equalising the 2 energies,
55.08 = mCΔT
ΔT = 55.08 / mC
= 55.08 / 2.7 x 10^-3 x 234
= 87.17
Hence, the increase in temperature of the bullet in degrees C is 87.17
To learn more about specific heat capacity from the given link
https://brainly.com/question/21406849
#SPJ1
An airplane covers a straight-line distance of 8.13 km in 33.5 s, during which time it has a constant forward acceleration of 4.6 m/s2.1. what is the speed at the first begining of 33.5 s.2.what is the speed at the end of the 33.5 s.
Given data
*The given distance is s = 8.13 km = 8130 m
*The given time is t = 33.5 s
*The given acceleration is a = 4.6 m/s^2
(1)
The formula for the speed at the first beginning of 33.5 seconds is given as
[tex]s=ut+\frac{1}{2}at^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 8130=u(33.5)+\frac{1}{2}(4.6)(33.5)^2 \\ u=165.6\text{ m/s} \end{gathered}[/tex]Hence, the speed at the first beginning of 33.5 s is u = 165.6 m/s
(2)
The formula for the speed at the end of the 33.5 s is given as
[tex]v=u+at[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=(165.6)+(4.6)(33.5)_{} \\ =319.7\text{ m/s} \end{gathered}[/tex]Hence, the speed at the end of the 33.5 s is v = 319.7 m/s
A cyclist is freewheeling (not exerting additional force) down a 7 degree angle hill. The cyclists weight is 75N. What acceleration is the cyclist experiencing? I have to do the following:1. Draw a free body diagram2. Identify Givens and Unknowns3. Identify the Equations4. Set up the equation using the givens and unknowns5. Solve
The free body diagram in shown below:
From the diagram and the problem we have that:
• The weight and angle of the inclined plane are given.
,• The normal force and the components of the weight are unknown (this implies that the acceleration is unknown too); we also notice that the mass is not given the it is also an unknown.
We know that Newton's second law states that:
[tex]\vec{F}=m\vec{a}[/tex]where F is the resultant force and a is the acceleration. Since this is a vector equation we can decomposed it in two scalar equations (in this case we only need two scalar equations since the forces are coplanar), then we have:
[tex]\begin{gathered} Wx=ma_x \\ N-W_y=ma_y \end{gathered}[/tex]Since we don't expect the cyclist to move in the y direction (otherwise he will surely fall) the equations above would reduce to:
[tex]\begin{gathered} W_x=ma \\ N-W_y=0 \end{gathered}[/tex]From the first equation we can solve the acceleration, to do this we use the triangle to get the x-component of the weight:
[tex]\begin{gathered} W_x=ma \\ W\sin \theta=ma \\ a=\frac{W\sin \theta}{m} \end{gathered}[/tex]Since the weight is given but not the mass we use the fact that the weight is:
[tex]W=mg[/tex]to get the mass, then we have:
[tex]\begin{gathered} m=\frac{W}{g} \\ m=\frac{75}{9.8} \\ m=7.65 \end{gathered}[/tex]hence the mass of the cylcist is 7.65 kg.
Now that we have all the values we need we plug them in the expression for the acceleration:
[tex]\begin{gathered} a=\frac{75\sin 7}{7.65} \\ a=1.19 \end{gathered}[/tex]Therefore the acceleration of the cyclist is 1.19 meters per second per second.
Good morning I really need some help on this question!
Sound travels faster in water than air
Here, C is the correct option.