We can use the following formula:
[tex]q=m\cdot c\cdot(T2-T1)[/tex]Where:
q = heat = 1 kcal = 4184000J
c = specific heat (of silver) = 0.235 J °C /g
m = mass
T1 = Initial temperature = 10°C
T2 = Final temperature = 90°C
Therefore, solve for m:
[tex]\begin{gathered} m=\frac{q}{c(T2-T1)} \\ m=\frac{4184000}{0.235(90-10)} \\ m=222553.1915g \end{gathered}[/tex]Answer:
222553.1915g
A massless scaffold is held up by a wire at each end. The scaffold is 12 m long. A300-N box sits 4.0 m from the left end. What is the tension in each wire?1) left wire = 100 N; right wire = 200 N2) left wire = 200 N; right wire = 100 N3) left wire = 900 N; right wire = 2700 N4) left wire = 2700 N; right wire = 900 N
Free body diagram:
Given data:
Length of massless scaffold (end to end) L=12 m.
Weight of box m=300 N.
Length of massless scaffold (center to end) l=6 m.
As, the box sits 4.0 m from the left end, the distance of the box from the center of massless scaffold is given as,
[tex]\begin{gathered} r=6.0\text{ m}-4.0\text{ m} \\ =2.0\text{ m} \end{gathered}[/tex]Balancing force in y direction,
[tex]T_1+T_2=300\text{ N}\ldots(1)[/tex]The torque is given as,
[tex]\tau=perpendicular\text{ distance}\times force[/tex]Therefore, torque along the center of massless scaffold is given as,
[tex]\begin{gathered} \Sigma\tau=0 \\ l\times T_1+r\times(300\text{ N})-l\times\tau_2=0 \\ 6\times T_1+2\times(300\text{ N})-6\times T_2=0 \\ 6T_1+600\text{ N}-6T_2=0 \\ 6(T_1+100\text{ N}-T_2)=0 \\ T_1+100\text{ N}-T_2=0 \\ T_1-T_2=-100\text{ N}\ldots(2) \end{gathered}[/tex]Adding equation (1) and (2),
[tex]\begin{gathered} (T_1+T_2)+(T_1-T_2)=300\text{ N}-100\text{ N} \\ T_1+T_1+T_2-T_2=200\text{ N} \\ 2T_1=200\text{ N} \\ T_1=\frac{200\text{ N}}{2} \\ T_1=100\text{ N} \end{gathered}[/tex]Substituting T1 in equation (1) we get,
[tex]\begin{gathered} 100\text{ N}+T_2=300\text{ N} \\ T_2=300\text{ N}-100\text{ N} \\ T_2=200\text{ N} \end{gathered}[/tex]Therefore, tension in left wire is 100 N and tension on right wire is 200 N. Hence, option (1) is the correct choice.
A human heart found to beat seventy five times in a minute. Calculate the beat frequency?
[tex]{ \green{ \tt{f = \frac{number \: of \: beats}{time \: taken}}}} [/tex]
[tex]{ \green{ \tt{number \: of \: beats = 75}}}[/tex]
[tex]{ \green{ \tt{time \: taken =1 \: min \: = 60 \: sec}}}[/tex]
[tex]{ \red{ \sf{f = \frac{ \cancel{75^{3}}}{ \cancel{ 60_{4} }}}}}[/tex]
[tex]{ \blue{ \boxed{ \purple{ \sf{f = \frac{3}{4} = 1.2 {s}^{ - 1}}}}}} [/tex]
___________________________________
[tex]{ \blue{ \sf{T = \frac{1}{f}}}} [/tex]
[tex]{ \blue{ \sf{T = \frac{1}{ \purple{ \sf{1.2}}}}}} [/tex]
[tex]{ \boxed{ \red{ \sf{T = 0.8 \: S}}}}[/tex]
A solenoid is wound with 259 turns per cm. An outer layer of insulated wire with 51 turns per cm is wound over the first layer of wire. The inner coil carries a current of 7.577 A, and the outer coil carries a current of 21.68 A in the opposite direction. What is the magnitude of the magnetic field, in microTeslas, at the central axis ?
Answer tab
For this question, we'll first define a number, called the "linear density of the coil" which is exactly the one the exercise gives us, in a unit of turns per length. The magnetic field can be calculated as:
[tex]B=\mu_0in[/tex]Where n is the linear density.
In our case, as the coils carry current in opposite directions, the generated magnetic fields will be opposed, and we'll have:
[tex]B=B_{outer}-B_{inner}=\mu_0*(51*10^2*21.68-259*10^2*7.577)[/tex]Please note that we had to multiply by 10^2, in order to convert turns/cm to turns/m
Then, our final magnetic field will be:
[tex]|B|=0.107664T=107664\mu T[/tex]Our final answer is B=107664uT
A can sits on a vertical wooden fencepost 1.9 meters above the ground. Billy picks up a small rock, aims at an angle ϴ = 25⁰ above the horizontal and throws the rock, releasing it 1 m above the ground with an initial speed of v0 =10 m/s. Boom! He hits the can! How far away is the fencepost?
Given,
Height of the fencepost, h=1.9 m
Angle at which the rock was thrown, θ=25°
The height at which the rock was released, a=1 m
The initial speed of the rock, v₀=10 m/s
Referring to the diagram,
[tex]\tan \theta=\frac{h-a}{d}[/tex]On rearranging the above equation,
[tex]d=\frac{h-a}{\tan \theta}[/tex]On substituting the known values,
[tex]d=\frac{1.9-1}{\tan 25^0}=1.93\text{ m}[/tex]Therefore the fencepost is at a distance of 1.93 m
You yell down a very deep well and it takes 1.5 s for your echo to return. If the speed of sound is 340 m/s then how deep is the well ?
We know that
• The time of the echo to return is 1.5 seconds.
,• The speed of the sound is 340 m/s.
It's important to consider that the sound wave has a constant speed, that is, it doesn't change its velocity. Therefore, we have to use the relation
[tex]d=v\cdot t[/tex]Where t = 1.5 sec and v = 340 m/s. Let's find d
[tex]\begin{gathered} d=340m/s\cdot1.5\sec \\ d=510m \end{gathered}[/tex]Hence, the well is 510 meters deep.A wheel Was spinning at 2.8 rad/s. It took 3.2 seconds to stop completely. What is the acceleration of the wheel?
Given
The angular velocity is
[tex]\omega=2.8\text{ rad/s}[/tex]The time taken,
[tex]t=3.2s[/tex]To find
The acceleration of the wheel
Explanation
The acceleration is
[tex]\begin{gathered} \alpha=\frac{\omega}{t} \\ \Rightarrow\alpha=\frac{2.8}{3.2}=\frac{0.875rad}{s^2} \end{gathered}[/tex]Conclusion
The acceleration is
[tex]0.875\text{ rad/s}^2[/tex]How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer: ___________ J (round to the nearest whole number)
According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:
[tex]W=\Delta K[/tex]Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:
[tex]K=\frac{1}{2}mv^2[/tex]Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:
[tex]\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\ \\ \therefore K\approx532,346J \end{gathered}[/tex]Therefore, the answer is: 532,346 J.
A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30°angle to the horizontal. The force of kinetic friction exactly balances the component ofgravity down the plane so that the plate, once started, glides down at constant velocity. If90% of the mechanical energy of the system is absorbed by the aluminum, what is itstemperature increase at the bottom of the incline? (Specific heat for aluminum is 900J/kg⋅°C.) Why do I multiply 15 by sin30?
A scheme of a the given situation is shown below:
First, consider that the work over the plate is done only by the component of the weight parallel to the incline (due to the perpendicular component is balanced by the friction force), then, the work on the plate is:
W = m*g*d*sinθ
where,
m: mass = 0.2kg
d: length of the incline = 15m
g: gravitational acceleration constant = 9.8m/s^2
θ = 30
By replacing the previous values into the expression for W, you obtain:
W = (0.2 kg)(9.8 m/s^2)(15 m)sin(30)
W = 14.7 J
Now, take into account that the amount of heat absorbed by the aluminum plate is given by the following formula:
Q = m*c*ΔT
Q: heat
m: mass
c: specific heat
ΔT: change in tempetaure
Take into account that the 90% of the mechanical energy is absorbed by the plate, which means that 0.9 of the work is converted to absorbed heat by the plate.
Then, you can write:
0.9W = Q
0.9(14.7J) = Q
13.23J = Q
Replace the given expression for Q into the previous equation and solve for ΔT, as follow:
m*c*ΔT = 13.23 J
ΔT = 13.23J/(m*c)
Now, replace the values of m and c for aluminum and simplify:
ΔT = 13.23J/(0.2kg*900J/kg°C)
ΔT = 0.0735°C
Hence, the temperature increase at the bottom of the incline is approximately 0.07°C
For an object starting from rest and accelerating with constantacceleration, distance traveled is proportional to the square of thetime. If an object travels 2.0 furlongs in the first 2.0 s, how far willit travel in the first 4.0 s?
Since the object is accelerating with constant acceleration we can use the following formula for the position of the object:
[tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]where x0 is the initial position, v0 is the initial velocity, a is the acceleration and t is the time. In this case, the initial position and velocity are zero. Plugging the values given we have:
[tex]\begin{gathered} 2=\frac{1}{2}a(2)^2 \\ 2=2a \\ a=1 \end{gathered}[/tex]Hence, the acceleration of the object is 1 furlong per second per second.
Once we know the acceleration we can use the same formula to determine how far the object will travel in four seconds.
[tex]\begin{gathered} x=\frac{1}{2}(1)(4)^2 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Therefore, the object will travel 8 furlongs in four seconds.
12000 inches to yards
ANSWER
[tex]\begin{equation*} 333.33\text{ yds} \end{equation*}[/tex]EXPLANATION
We want to convert 12000 inches to yards.
To do this, divide the value in inches by 36:
[tex]\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}[/tex]That is the answer.
Water flows through a pipe diameter of 8.000 cm at 49.0 m/min. Find the flow rate in m3/min
We are asked to determine the volumetric flow rate through a pipe of diameter 8.000 cm. To do that we will use the following formula:
[tex]R=Av[/tex]Where:
[tex]\begin{gathered} R=\text{ volumetric flow rate} \\ A=\text{ cross-area of the pipe} \\ v=\text{ velocity of the flow} \end{gathered}[/tex]The cross-area of the pipe is the area of a circle and is given by:
[tex]A=\frac{\pi D^2}{4}[/tex]Where:
[tex]\begin{gathered} A=\text{ cross-area} \\ D=\text{ diameter} \end{gathered}[/tex]Before we determine the area we will convert the diameter from cm to meters using the following conversion factor:
[tex]100cm=1m[/tex]Multiplying by the conversion factor we get:
[tex]8.000cm\times\frac{1m}{100cm}=0.080m[/tex]Now, we plug in the value in the formula for the area:
[tex]A=\frac{\pi(0.080m)^2}{4}[/tex]Solving the operations:
[tex]A=0.005m^2[/tex]Now, we plug in the values of area and velocity in the formula or the volumetric flow rate:
[tex]R=(0.005m^2)(49.0\frac{m}{\min })[/tex]Solving the operations:
[tex]R=0.246\frac{m^3}{min}[/tex]Therefore, the flow rate is 0.246 cubic meters per minute.
I think this is all statements are true but I just want to make sure
Given that a bug flies into the windshield of a car going. Let's select the correct statements.
According the NEwton's third law, the force exerted on the bug by the car is equal to the force extered on the car by the bug.
To determine the acceleration, we have:
[tex]a=\frac{F}{m}[/tex]Where:
F is the force
m is the mass
The mass of the car will be greater than the mass of the bug.
Since the mass of the car is greater than the mass of the bug and they have the same force, we can say the acceleration of the bug is greater than the acceleration of the car.
Statement B is correct.
The force of impact is the same for both according to Newton's third Law.
Both the car and the bug deliver the same magnitude of impulse on each other.
Therefore, all statements are correct.
ANSWER:
All statements are true.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
A feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. Which graph shows the total mechanical energy of the bowling ball as it falls?
The total mechanical energy of the bowling ball and the feather is shown by the graph in option D
What is the total mechanical energy?We know that the mechanical energy is the energy that is possessed by a body by virtue of its motion or by virtue of its staying at a place. Thus mechanical energy is possessed by an object that is moving or by an object that is at rest.
In this case, we have a feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. We know that the mechanical energy of the two objects must be constant. This is because, the potential energy of the feathers and the ball at a height is converted to kinetic energy as the two objects begin to move.
The graph that would show the total energy must be one in which the energy axis of the graph is constant as we see in option D.
Learn more about mechanical energy:https://brainly.com/question/13552918
#SPJ1
Answer:
the answer is D
Explanation:
A 10 gram ball is rolling at 3 m/s. Calculate its kinetic energy.
ANSWER:
0.045 joules
STEP-BY-STEP EXPLANATION:
Given:
mass (m) = 10 g = 0.01 kg
velocity (v) = 3 m/s
The kinetic energy is given by the following formula:
[tex]K_E=\frac{1}{2}mv^2[/tex]We replacing:
[tex]\begin{gathered} K_E=\frac{1}{2}\cdot0.01\cdot3^2 \\ K_E=0.045\text{ J} \end{gathered}[/tex]The kinetic energy is 0.045 joules.
If a rock has 376 J of potential energy when it’s held 10.1 m above the ground what is its mass? Round to the nearest tenth
Answer: 3.8 kg
Explanation:
The formula for calculating potential energy is expressed as
Potential energy = mgh
where
m is the mass of the object
g is the acceleration due to gravity
h is the height above the ground
From the information given,
Potential energy = 376
g = 9.8 m/s^2
h = 10.1
Thus,
376 = m x 9.81 x 10.1
376 = 99.081m
m = 376/99.081
m = 3.8 kg
The mass is 3.8 kg
If the man and woman are taken to a planet where the acceleration due to gravity is twice that of earth repeat the woman mass was 25kg on earth and the man was 300N on another planet
The mass of the woman is 25 kg because the mass is constant.
The mass of the man can be found using the formula: W = mg, where g is double Earth's gravity.
[tex]\begin{gathered} m=\frac{W}{g} \\ m=\frac{300N}{2\cdot9.8\cdot\frac{m}{s^2}} \\ m=\frac{300}{19.6}kg \\ m\approx15.3\operatorname{kg} \end{gathered}[/tex]The mass of the man is 15.3 kg.
The weight of the man on Earth can be found with the same formula but using Earth's gravity.
[tex]\begin{gathered} W=15.3\operatorname{kg}\cdot9.8\cdot\frac{m}{s^2} \\ W=149.94N \end{gathered}[/tex]The weight of the man on Earth is 149.94 N.
At last, the weight of the woman on Earth can be found using the same method before.
[tex]\begin{gathered} W=25\operatorname{kg}\cdot9.8\cdot\frac{m}{s^2} \\ W=245N \end{gathered}[/tex]The weight of the woman on Earth is 245N.
This question is based on Oscillations and waves. I tried it for days and I just couldn't get it right.
ANSWER:
The maximun velocity is 16.07 m/s
At x = 0.26
The velocity is 8.36 m/s
The accelearion is 286.67 m/s^2
The resorting force is 86 N
STEP-BY-STEP EXPLANATION:
Given:
k = 310 N / m
Max distance = 0.5 m
Mass of block = 0.3 kg
Max velocity:
Using conservation of energy:
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v^2=\frac{kx^2}{m} \\ \text{ replacing} \\ v^2=\frac{310\cdot0.5^2}{0.3} \\ v=\sqrt[]{258.33} \\ v=16.07\text{ m/s} \end{gathered}[/tex]At x = 0.26 m:
[tex]\begin{gathered} v^2=\frac{kx^2}{m} \\ v^2=\frac{310\cdot0.26^2}{0.3} \\ v=\sqrt[]{69.85} \\ v=8.36\text{ m/s} \end{gathered}[/tex]Acceleration:
[tex]\begin{gathered} F=k\cdot x \\ F=m\cdot a \\ \text{ therefore} \\ m\cdot a=k\cdot x \\ a=\frac{k\cdot x}{m} \\ \text{ replacing} \\ a=\frac{310\cdot0.26}{0.3} \\ a=286.67\text{ }\frac{m}{s^2} \end{gathered}[/tex]The resorting force:
[tex]\begin{gathered} F=m\cdot a \\ \text{ replacing} \\ F=0.3\cdot286.67 \\ F=86\text{ N} \end{gathered}[/tex]Help me with number 1 I’m very lost. I just need the equation. No explanation I’m stuck on 1 part.
Given data:
* The value of angular velocity is,
[tex]\omega=1.3\text{ rad/s}[/tex]Solution:
(a). The time period of the oscillation in terms of the angular velocity is,
[tex]T=\frac{2\pi}{\omega}[/tex]Substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi}{1.3} \\ T=4.83\text{ s} \end{gathered}[/tex]Thus, the time period of oscillation is 4.83 s.
(b). The frequency of the oscillation in terms of the time period is,
[tex]undefined[/tex]A radioactive tracer has a half-life of two hours how much of a 2500 g sample will be available after 18 hours?
ANSWER:
4.88 grams
STEP-BY-STEP EXPLANATION:
We must first calculate how many half-life there are in 18 hours, knowing that each half-life takes 2 hours.
[tex]\frac{18}{2}=9\text{ half-life}[/tex]Now, knowing this, we can calculate the number of grams applying 9 times the half-life, like this:
[tex]\frac{2500}{2^9}=4.88\text{ g}[/tex]Which means that after 18 hours there are 4.88 grams
A concave mirror of focal length 10 cm forms an upright and diminished image of a real object placed at a distance of 5 cm from the mirror. Is this true or false?
Given,
The focal length of the concave mirror, f=10 cm
The object distance, o=5 cm
From the mirror formula, we have,
[tex]\frac{1}{f}=\frac{1}{o}+\frac{1}{i}[/tex]Where i is the image distance.
On substituting the known formula,
[tex]\begin{gathered} \frac{1}{10}=\frac{1}{5}+\frac{1}{i} \\ \Rightarrow\frac{1}{i}=\frac{1}{10}-\frac{1}{5} \\ =-\frac{1}{10} \\ \Rightarrow i=-10\text{ cm} \end{gathered}[/tex]And the magnification is given by,
[tex]m=\frac{-i}{o}[/tex]On substituting the known values,
[tex]\begin{gathered} m=\frac{-(-10)}{5} \\ =2 \end{gathered}[/tex]Thus the object is not diminished, it is magnified.
Thus the given statement is false.
explain how the intensity of the UV light vaires across the Earth
Some factors determine the amount of UV radiation that reach certain part of Earth's surface. They are listed and briefly explained below:
• Cloud coverage. Water molecules on clouds scatter the radiation, hence the more clouds the less UV radiation.
,• Ozone. Similarly as cloud coverage, the more concentration of ozone the less UV radiation that reaches the surface of the Earth.
,• Angle of incidence. If the angle of incidence of the UV light is oblique the light will spread in a wider area, and hence the intensity is spread across this area.
,• Aerosols. The molecules of aerosols also scatter the UV light.
,• Elevation. The more the elevation the greater the amount of UV light.
If you have a convex lens whose focal length is 10.0 cm, where would you place an object in order to produce an image that is virtual?
When an object is placed between first focus and optical center of a convex lens then virtual image is produced.
Here , focal length is 10.0 cm . So object distance should be less than 10.0 cm
Final answer is : between focus and optical center of the lens
Objects with masses M1=12.0 kg and M2= 7.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, M2 falls 1.00m in 1.33s, determine the coefficient of Kinect friction between M1 and the table
We are asked to determine the coefficient of friction in the sysmtem. First we will do a free body diagram of the first mass, like this:
Where:
[tex]\begin{gathered} N=\text{ Normal Force} \\ m_1=\text{ mass} \\ g=\text{ acceleration of gravity} \\ T=\text{ tension} \\ F_f=\text{friction force} \end{gathered}[/tex]Now, we add the forces in the horizontal direction, we get:
[tex]T-F_f=m_1a[/tex]Now, to determine the friction force we need to use the following relationship:
[tex]F_f=\mu N[/tex]To determine the normal force we will add the forces in the vertical direction. Since there is no movement in the vertical direction this sum must add up to zero:
[tex]\begin{gathered} N-m_1g=0 \\ N=m_1g \end{gathered}[/tex]Now, we substitute the value of the normal force in the equation for the friction force:
[tex]F_f=\mu m_1g[/tex]Now, we substitute the friction force in the sum of horizontal forces:
[tex]T-\mu m_1g=m_1a[/tex]Now, we turn our attention to the second mass. We add the forces in the vertical direction:
[tex]m_2g-T=m_2a[/tex]Now, since the acceleration "a" and the tension "T" is the same for both masses we will solve fot "T" in the sum of forces for the second mass, like this:
[tex]m_2g-m_2a=T[/tex]Now, we substitute the value in the sum of forces of the first mass, we get:
[tex]m_2g-m_2a-\mu m_1g=m_1a[/tex]Now, we solve for the coefficient of friction. To do that we will subtract "m2g" to both sides:
[tex]-m_2a-\mu m_1g=m_1a-m_2g[/tex]Now, we add "m2a" to both sides:
[tex]-\mu m_1g=m_1a-m_2g+m_2a[/tex]Now, we divide both sides by "-m1g":
[tex]\mu=\frac{m_1a-m_2g+m_2a}{-m_1g}[/tex]We have gotten an expression for the coefficient of friction but we need to determine the acceleration of the system. To do that we will use the fact that the second mass moves 1 meter in 1.33 seconds. Assuming constant acceleration we can use the following equation of motion:
[tex]y=v_0t+\frac{1}{2}at^2[/tex]Where:
[tex]\begin{gathered} y=\text{ distance} \\ v_0=\text{ initial velocity} \\ t=\text{ time} \\ a=\text{ acceleration} \end{gathered}[/tex]Since the mass starts from rest we have that the initial velocity is zero, therefore:
[tex]y=\frac{1}{2}at^2[/tex]Now, we solve for the acceleration. First, we multiply both sides by 2:
[tex]2y=at^2[/tex]Now, we divide both sides by the time squared:
[tex]\frac{2y}{t^2}=a[/tex]Now, we plug in the values:
[tex]\frac{2(1m)}{(1.33s)^2}=a[/tex]Solving the operations:
[tex]1.13\frac{m}{s^2}=a[/tex]Now, we substitute the values in the equation for the coefficient of friction:
[tex]\mu=\frac{(12kg)(1.13\frac{m}{s^2})-(7kg)(9.8\frac{m}{s^2})+(7kg)(1.13\frac{m}{s^2})}{-(12kg)(9.8\frac{m}{s^2})}[/tex]Solving the operations we get:
[tex]0.4=\mu[/tex]Therefore, the coefficient of friction is 0.4
1 pts
An coconut with a mass of 2 kg and a feather with a mass of 0.01 kg fall from a tree through the air to the ground
below, both eventually reaching terminal velocity. At terminal velocity, the amount of air-rresistance force what is the answer ? A grather on the coconut B the same on each C grather on the feather
Take into account that air-resistance force is greater against bodies with lower densities.
In this case, the feather has a lower density than the coconut, then, you can conclude that air-resistance force is greater on the feather (option C).
What is the relationship. Stern average kinetic energy of a gas and it’s temperature?
Average kinetic energy is directly proportional to its temperature-
E= f/2 N k T
E= energy
T= temperature
A truck can travel at 100 km/hr How long would it take to drive 900km?
Given:
The speed of the truck is,
[tex]v=100\text{ km/hr}[/tex]The distance is,
[tex]s=900\text{ km}[/tex]The time to drive this distance is,
[tex]t=\frac{s}{d}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\frac{900}{100} \\ =9\text{ hrs} \end{gathered}[/tex]Hence, the time is 9 hrs.
I need help with this table pleasecalculate relative density of steel. Use table 3
Take into account that the relative density is given by:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}[/tex]where ρ, in this case, is the density of the steel and ρwater is the density of water (1000 kg/m^3).
The density of the steel is:
[tex]\rho=\frac{\text{mass}}{\text{volume}}[/tex]Based on table 3, you have:
mass = 50.7 g = 0.0507 kg
volume = 0.0000063 m^3
[tex]\rho=\frac{0.0507kg}{0.0000063m^3}\approx8047.62\frac{kg}{m^3}[/tex]Then, for the relative density you obtain:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}=\frac{8047.62\frac{kg}{m^3}}{1000\frac{kg}{m^3}}\approx8.048[/tex]Hence, the relative density of steel is 8.048
I need help with #2 it off s for practice
First, find the acceleration using Newton's Second Law.
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \end{gathered}[/tex]Where F = 8.10x10^5 N and m = 1.40x10^7 kg.
[tex]a=\frac{8.10\times10^5N}{1.40\times10^7\operatorname{kg}}=5.79\times10^{-2}\cdot\frac{m}{s^2}[/tex]Then, use a formula that relates acceleration, initial velocity, final velocity, and time.
[tex]v_f=v_0+at[/tex]Solve for t because the problem is asking to find the time.
[tex]\begin{gathered} v_f-v_0=at_{} \\ t=\frac{v_f-v_0}{a} \end{gathered}[/tex]Where vf = 64 km/h, v0 = 0, and a = 5.79x10^-2 m/s^2. Before we continue, we need to transform the final velocity to m/s.
[tex]v_f=64\cdot\frac{km}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600\sec}=17.78\cdot\frac{m}{s}[/tex]Once we have the velocity transformed, we are able to find t.
[tex]\begin{gathered} t=\frac{17.78\cdot\frac{m}{s}-0}{5.79\times10^{-2}\cdot\frac{m}{s^2}} \\ t=3.07\times10^2\sec \\ t=307\sec \end{gathered}[/tex]But, the answer must be in minutes.
[tex]t=307\sec \cdot\frac{1\min}{60\sec}=5.12\min [/tex]Therefore, it takes 5.12 minutes.
You wish to lift a 720N crate of bricks to the 3rd floor of a building in a construction site. The 3rd floor is 16m high. How much work will that require?
ANSWER
[tex]11,520J[/tex]EXPLANATION
Parameters given:
Weight (force), F = 720N
Height (distance), d = 16m
To find the work required to lift the crate to the 3rd floor, we have to find the product of the force (weight of the crate) and the distance it will be lifted.
Therefore, we have:
[tex]\begin{gathered} W=F\cdot d \\ W=720\cdot16 \\ W=11,520J \end{gathered}[/tex]That is the work that it will require.
I was told it was 5.886 J by another tutor on here but that was incorrect so just trying again
The potential energy of gravity is given by:
[tex]\begin{gathered} E=mgh \\ where: \\ m=0.3 \\ h=2 \\ g=9.8 \\ so: \\ E=0.3\cdot2\cdot9.8 \\ E\approx5.9J \end{gathered}[/tex]Answer:
5.9 J