In order to determine the speed of the Earth, proceed as follow:
Consider that the centripetal force must be equal to the gravitational force between the Earth and the Sun (because guarantees the stability of the system):
[tex]F_g=F_c[/tex]Fg is the gravitational force and Fc the centripetal force. The expressions for each of these forces are:
[tex]\begin{gathered} F_g=\text{G}\frac{\text{mM}}{r^2} \\ F_c=ma_c=m\frac{v^2}{r} \end{gathered}[/tex]where,
G: Cavendish's constant = 6.67*10^-11 Nm^2/kg^2
m: Earth's mass = 5.97*10^24 kg
M: Sun's mass = 1.99*10^30kg
v: speed of Earth around the Sun = ?
r: distance between the center of mass of Earth and Sun = 1.49*10^8km = 1.49*10^11 m
Equal the expressions for Fg and Fc, solve for v, replace the previous values of the parameters and simplify:
[tex]\begin{gathered} \text{G}\frac{\text{mM}}{r^2}=m\frac{v^2}{r} \\ v^{}=\sqrt[]{\frac{GM}{r}} \\ v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(1.99\cdot10^{30}kg)}{1.49\cdot10^{11}m}} \\ v\approx29846.7\frac{m}{s} \end{gathered}[/tex]Hence, the speed of the Earth around the Sun is approximately 29846.7m/s
johns mass is 92.0kg on the earth, what is his mass on mars where g= 3.72m/s^2
The mass is a fundamental property of all matter.
Mass is not the same as weight, the weight depends on the gravity while the mass does not. Therefore the John's mass is the same in mars.
What is the index of refraction for a medium where light has a velocity of 2.75 x10³ m/s?0.9176.751.090.250
n = c/v, where:
n: index of refraction
c: speed of light in a vacuum (3*10^8 m/s)
v: speed of light in medium
n = (3*10^8)/(2.75*10^8)
n = 3/2.75
n = 1.09
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
Learn more about acceleration here: https://brainly.com/question/14344386
#SPJ1
A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 219 m, and the car completes the turn in 32.0 s.
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors î and ĵ.
(b) Determine the car's average speed.
(c) Determine its average acceleration during the 32.0-s interval.
a.) the acceleration when the car is at B located at an angle of 35.0° is 2.689i -0.42818j
b.) the car's average speed is v= 6.84375 m/s
c.) average acceleration during the 32.0-s interval is (−0.181 i+0.181 j)m/s²
What is acceleration?Acceleration is described as the rate of change of the velocity of an object with respect to time.
(a) The car’s speed around the curve is found from
v= 219/32.0
v= 6.84375 m/s
This is the answer to part (b) of this problem. We calculate the radius of the curve from
(1/4) X 2πr = 219 m
which gives r = 139.4 m
The car’s acceleration at point B is then
ar = (V²/r ) towards the center
= ( 6.84375)² / 139.4 at 35.0° north of west
= (2.9761 m/s²)(cos 35.0)(-i) + (sin 35.0j)
= -2.689i -0.42818j
(b) From part (a), v= 6.84375 m/s
(c) We find the average acceleration from
A avg = (Vf - Vi)/ change in t
A avg = ( 6.84375 j - 6.84375 i ) / 32.0 s
A avg = (−0.181 i+0.181 j)m/s²
Learn more about acceleration at: https://brainly.com/question/605631
#SPJ1
Do you think we can see the planets without telescope? explain your answer
Answer:
No, how do you tell that?
Telescope is an example of technology that use on seeing the object on far place try to do that, you acn see a planet like small pieceof like an tungaw
Accomplishment of SPTA Officers and all parents this 1st Quarter.. Enjoy watching..Mabbalo
Explanation:
HOPE IT HELP TO YOUA) A recipe calls for 5.0 qt of milk. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.B) A gas can holds 2.0 gal of gasoline. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.
A.
[tex]\begin{gathered} 1qt=946.353cm^3 \\ 5qt=\text{?} \\ \text{quantity}=946.353\times5 \\ \text{quantity}=4731.765cm^3 \end{gathered}[/tex]B.
[tex]\begin{gathered} 1\text{ gal=}3785.41\text{ }cm^3 \\ 2\text{ gal}=\text{?} \\ \text{quantity}=2\times3785.41 \\ \text{quantity}=7570.82cm^3 \end{gathered}[/tex]How many moles of a gas sample are in a 5.0 L container at 251 K and 370 kPa?(The gas constant is8.31L kPamol K)Round your answer to one decimal place and enter the number only with no units.
Given:
The volume of the gas, V=5.0 L
The temperature of the gas, T=251 K
The pressure of the gas, P=370 kPa
The gas constant, R=8.31 L kPa/(mol K)
To find:
The moles of the gas sample.
Explanation:
From the ideal gas equation,
[tex]PV=\text{nRT}[/tex]Where n is the moles of the gas present.
On substituting the known values,
[tex]\begin{gathered} 370\times5.0=n8.31\times251 \\ \Rightarrow n=\frac{370\times5.0}{8.31\times251} \\ =0.9\text{ mol} \end{gathered}[/tex]Final answer:
The moles of the gas present in the sample is 0.9 mol
A type of wave that is a combination of a transverse and a longitudinal wave is called aQuestion 17 options:Slinky waveSurface waveSound waveLight wave
The type of wave that is a combination of longitudinal and Transverse wave is called Surface wave
calculate how much heat is released when 500g of platinum is cooled from 250.0K to 240.0K
Answer:
Q = 665 J
Explanation:
Given:
m = 500 g = 0.5 kg
T₁ = 250.0 K
T₂ = 240.0 K
c = 133 J / ( kg*⁰K) - Specific heat capacity of platinum
___________
Q - ?
Heat is released:
Q = c*m*( T₁ - T₂) = 133*0.5*(250.0 - 240.0) =
= 133*0.5*10 ≈ 665 J
A football player runs from his own goal line to the opposing team's goal line, returning to his twenty-yard line, all in 27.0 s. Calculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.)HINTApply the definitions of average speed and average velocity.Click the hint button again to remove this hint.(a) Calculate his average speed. ____yards/s(b) Calculate the magnitude of his average velocity. ____yards/s
time = 27 s
d = 20 yard
a) speed = distance / time
d1 = 100 yard ( own goal to opposing team's goal line)
d2 = 80 yard ( returning to 20 yard line)
d= d1+d2
d= 100 + 80 = 180 yards
Speed = 180 y / 27s = 6.667 y/s
b) velocity = displacement / time
d = 100 - 80 = 20 y
Velocity = 20 / 27 = 0.74 y/s
Which has more kinetic energy: a 0.0014-kg bullet traveling at 397 m/s or a 5.9 107-kg ocean liner traveling at 13 m/s (25 knots)?
the bullet has greater kinetic energy
the ocean liner has greater kinetic energy
Justify your answer.
Ek-bullet =
J
Ek-ocean liner =
J
Answer:
Ocean Liner because its mass is 5000 times more and the bullet's velocity squared is only 1600 times more. This means that the kinetic energy of the ocean liner will be roughly 3 times greater.
Explanation:
The kinetic energy of an object is dependent on mass and velocity.
E = ( 1 / 2 )mv²;
Energy is measured in Joules which is equal to kilogram metres squared per second squared.
1J = ( kg )( m² )( s⁻² ) or 1J = ( kg )( m² ) / ( s² )
We can substitute the mass and velocity directly into the equation because the question gives the values in metres and kilograms.
BULLET:
E = ( 1 / 2 )( 0.0014kg )( 397m / s )²;
E = ( 0.0007kg )( 157609m² )( s⁻² );
E = 110.3263kgm²s⁻²;
E = 110.3263J;
I will just skip some steps because you get the idea.
OCEAN LINER:
E = ( 1 / 2 )( 5.9107kg )( 13m / s )²;
E = 499.45415J;
Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The mass of the Bart is m_1 = 32.4 kg.
* The mass of the Milhouse is m_2 = 27.6 kg.
* The distance between the Millhouse and Bart is d = 4 m.
Solution:
To balance the seesaw, the net moment about the center should be zero.
The diagrammatic representation of the given system is,
The distance between the Bart and Milhouse can be written as,
[tex]\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}[/tex]where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,
Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.
Thus, the net moment about the center is,
[tex]M=m_1d_1-m_2d_2[/tex]Substituting the known values,
[tex]\begin{gathered} 0=32.4\times d_1-27.6\times(4-d_1) \\ 0=32.4\times d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} 60d_1=110.4 \\ d_1=\frac{110.4}{60} \\ d_1=1.84\text{ m} \end{gathered}[/tex]Thus, the distance of the Bart from the center is 1.84 meters.
Forces that are equal in magnitude
but opposite in direction will:
Answer:
The two forces equal in magnitude but acting opposite in direction on a body are called balanced forces
Explanation:
A bunny and a tortoise start a race from rest. The bunny accelerates at a rate "a" for a time "t" until it reaches itsmaximum speed (vb) after traveling a distance d_b. The tortoise accelerates at a rate one fourth as great as the bunny andtakes three times as long to reach its maximum speed of vt after traveling a distance d_t.a) What is the maximum speed of the tortoise? Answer in terms of vbb)How far did the tortoise travel? Answer in terms of db
We are given that a Bunny starts a race from rest, this means that its initial speed is zero:
[tex]v_{b0}=0,(1)[/tex]We are also given that it accelerates at a rate "a" for a time "t" and reaches its maximum speed. We can use the following equation of motion to relate the final speed with its acceleration:
[tex]v_b=at,(2)[/tex]We are also given that it travels a distance "d". We can use the following equation of motion to relate the distance travel with the velocity and the acceleration:
[tex]2ad_b=v^2_b,(3)[/tex]Now. We turn our attention to the turtle. We are told that its acceleration is one-fourth of the acceleration of the bunny, therefore, we have:
[tex]a_t=\frac{1}{4}a[/tex]And it also says that it takes 3 times as long to reach its maximum speed, therefore the time of the turtle is:
[tex]t_t=3t[/tex]Now we use the equation of motion that relates velocity with time and acceleration for turtle:
[tex]v_t=a_tt_t[/tex]Since we are required to express this velocity in terms of the final velocity of the bunny we replace the values of acceleration and time of the turtle that we determined previously:
[tex]v_t=(\frac{1}{4}a)(3t)[/tex]Rearranging terms:
[tex]v_t=\frac{3}{4}at[/tex]From the equation of motion from the bunny we know that the product of acceleration and time of the bunny equals its final velocity , therefore:
[tex]v_t=\frac{3}{4}v_b[/tex]And thus we have determined the final velocity of the turtle in terms of the final velocity of the bunny.
Now we are told to determine the final velocity in terms of the distance traveled by the bunny. First, we will use the equation of motion that relates the distance, the acceleration, and the time with the final velocity:
[tex]2a_td_t=v^2_t[/tex]Now we replace the value of the acceleration:
[tex]2(\frac{1}{4}a)d_t=v^2_t[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=v^2_t[/tex]Now we use the relationship between velocities we determined in part A:
[tex]\frac{1}{2}ad_t=(\frac{3}{4}v_b)^2[/tex]Simplifying:
[tex]\frac{1}{2}ad_t=\frac{9}{16}v^2_b[/tex]From equation (3) we can replace the value of the velocity of the bunny:
[tex]\frac{1}{2}ad_t=\frac{9}{16}\times2ad_b[/tex]We can cancel out the acceleration:
[tex]\frac{1}{2}d_t=\frac{9}{16}\times2d_b[/tex]Now we multiply both sides by 2:
[tex]d_t=2\times\frac{9}{16}\times2d_b[/tex]Simplifying:
[tex]d_t=\frac{9}{4}d_b[/tex]And thus we have found the distance traveled by the turtle in terms of the distance traveld by the bunny.
I need help with this question There is 4 answersa)b)c)d)
Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
[tex]\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
[tex]\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
[tex]\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
[tex]\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system, in this case, is 143.5 kg m/s
Mechanical energy is the form of energy associated with the ,or of an object.
Mechanical energy is the form of energy associated with the motion or position of an object.
Explanation:Mechanical energy can be defined as the energy possessed by an object as a result of its motion or position.
Mechanical energy is divided into two.
Kinetic energy and potential energy
Kinetic energy is the energy possessed by an object as a result of its motion
Potential energy is the energy possessed by an object as a result of its position.
Therefore, we can conclude that mechanical energy is the form of energy associated with the motion or position of an object.
Object a attracts objects be the gravitational force of 10 N from a given distance the distance between the two objects is doubled what is the new force of attraction between
We are given that two objects are being attracted by a gravitational force between each other of 10N. The gravitational force between two masses is given by the following equation:
[tex]F_g=G\frac{m_Am_B}{r^2}[/tex]Where:
[tex]\begin{gathered} F_g=\text{ Gravitational force} \\ m=\text{mass} \\ G=\text{ Gravitational constant} \\ r=\text{ distance between the masses} \end{gathered}[/tex]Replacing the given values for the 10N force:
[tex]10=G\frac{m_Am_B}{r^2_1}[/tex]Where:
[tex]r_1=\text{ initial distance}[/tex]Now we will solve for the product of the masses and the gravitational constant by multiplying both sides by the distance squared:
[tex]10r^2_1=Gm_Am_B[/tex]Now, the product of the masses and the gravitational constant won't change if we double the distance, therefore, if we apply the equation for the gravitational force for the new distance we get:
[tex]F_{g2}=G\frac{m_Am_B}{r^2_2}[/tex]We can replace the value we determined earlier:
[tex]F_{g2}=\frac{10r^2_1}{r^2_2}[/tex]Since the distance is double, we have:
[tex]r_2=2r_1[/tex]Replacing in the previous equation:
[tex]F_{g2}=\frac{10r^2_1}{(2r_1)^2}[/tex]Solving the square:
[tex]F_{g2}=\frac{10r^2_1}{4r^2_1}[/tex]Now we can cancel out the distances squared:
[tex]F_{g2}=\frac{10}{4}[/tex]Solving the operation:
[tex]F_{g2}=2.5[/tex]Therefore, doubling the distance the new gravitational force is 2.5N.
If you have 100.g of a radioactive isotope with a half-life of 10. years, how much of the isotopewill you have left after 20. years?
Use the radioactive decay formula using t_0 as the half life:
[tex]A=A_0\cdot2^{-t/t_0}[/tex]Substitute the initial amount of radioactive element A_0=100g, the half life t_0=10y and the time period t=20y to find the remaining amount after 20 years:
[tex]\begin{gathered} A=100g\times2^{-20y/10y} \\ =100g\times2^{-2} \\ =100g\times\frac{1}{4} \\ =25g \end{gathered}[/tex]Therefore, after 20 years, there are 25 grams left.
A 75 kg criminal wants to escape from the 5th story window of the jail, 24 m above the ground. He has a rope, which can only support a tension force of 650 N.
a. What is the maximum acceleration he can slide down without breaking his "rope?"
Answer:
a = 1.1 m/s²
Explanation:
Given:
P = 650 N
m = 75 kg
g = 9.8 m/s²
____________
a - ?
Forse:
P = m*(g - a)
650 = 75*(10 - a)
10 - a = 650 /75
Acceleration:
a = 9.8 - 650 / 75 = 9.8 - 8.7 = 1.1 m/c²
A star of mass 3.0e30 kg is moving in a circle of radius 1.0e12 metres, with a period of 100 years. This is due to the gravity of a second unseen object. what is the mass of the unseen object in kg
The mass of the unseen object is 1.21 x 10²⁴ kg.
What is the centripetal acceleration of the star?
The centripetal acceleration of the star is calculated as follows;
a = v²/r
where;
v is the linear speed of the starr is the radiusv = 2πr/T
where;
T is period of the motion = 100 years = 3.154 x 10⁹ sv = (2π x 1 x 10¹²) / (3.154 x 10⁹)
v = 1,992.1 m/s
a = (1992.1)² / ( 1 x 10¹²)
a = 3.97 x 10⁻⁶ m/s²
The centripetal force of the star is calculated as follows;
F = ma
F = (3 x 10³⁰) x ( 3.97 x 10⁻⁶)
F = 1.19 x 10²⁵ N
The mass of the unseen object is calculated as follows;
F = mg
m = F/g
where;
g is the acceleration due to gravitym = (1.19 x 10²⁵) / (9.8)
m = 1.21 x 10²⁴ kg
Learn more about centripetal force here: https://brainly.com/question/898360
#SPJ1
A marble was thrown vertically up the air and came back down to the samelevel in 1.6 s. (using g = 9.81 ms?).(a) What is it initial velocity?(b) What would be the highest height it can reach?(c) How long would it take to reach the highest point?
ANSWER:
(a) 15.7 m/s
(b) 12.56 m
(c) 0.8 sec
STEP-BY-STEP EXPLANATION:
We can calculate the initial velocity using the following formula:
(a)
[tex]\begin{gathered} v=u+a\cdot t \\ v=0 \\ \text{therefore:} \\ u=-a\cdot t \\ \text{ replacing} \\ u=-(-9.81\cdot1.6) \\ u=15.7\text{ m/s} \end{gathered}[/tex](b)
We calculate the height in this way:
[tex]\begin{gathered} h=\frac{(-u)^2}{2g} \\ \text{ replacing} \\ h=\frac{(-15.7)^2}{2\cdot9.81} \\ h=12.56\text{ m} \end{gathered}[/tex](c)The time to reach this height would be half the time it takes for the entire journey, that is, 0.8 seconds
Compute, in centimeters and in meters, the height of a basketball player who is 6 ft 1 in. tall into cm and m
We will have that:
6ft 1in is:
[tex]6ft(\frac{12in}{1ft})+1in=73in[/tex]Then:
[tex]73in(\frac{2.54cm}{1in})=185.42cm[/tex]So, the solution is both meters and cm are respectively.
1.8542 m and 185.42 cm.
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock?
The net force on the rock is determined as 76.2 N.
What is the net force on the rock?The net force on the rock is obtained by subtracting the weight of the rock from the input force on the lever.
F(net) = Input force - weight of the rock
The weight of the rock is calculated as follows;
W = mg
where;
m is mass of the rock = 6 kgg is acceleration due to gravity = 9.8 m/s²W = 6 x 9.8
W = 58.8 N
F(net) = 135 N - 58.8 N
F(net) = 76.2 N
Learn more about net force here: https://brainly.com/question/14361879
#SPJ1
The complete question is below:
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock? if the mass of the rock is 6kg.
Will mark as brainlist
Which of the following best represents R= A - B ?
Please help, it’s due soon!
Option C is the best representation for the resultant vector representing the resultant R=A+B because the result is represented by the closing side of the triangle, so it is the correct answer.
What exactly is a vector quantity?A measure with both magnitude and direction It is typically represented by an arrow with the same direction as the quantity and a length proportional to the magnitude of the quantity. A vector, while including magnitude and demand, does not have a position. Vector quantities are physical quantities that are distinguished by the presence of both magnitude and direction. Displacement, force, torque, momentum, acceleration, velocity, and so on. Vector quantities are physical quantities that have distinct magnitude and direction definitions. For example, suppose a boy is riding a bike at 30 km/hr in the north-east direction. In nature, vectors have velocity, acceleration, force, electromagnetic fields, and importance. (Weight is the force produced by gravity's acceleration acting on a mass.) A Scalar is a quantity or phenomenon that has only magnitude and no specific direction.To learn more about vector quantity, refer to:
https://brainly.com/question/2912049
#SPJ13
Sheena can row a boat at 2.90 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 2.00 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.What is her speed with respect to the starting point on the bank?How long does it take her to cross the river?How far upstream or downstream from her starting point will she reach the opposite bank? If upstream, enter a positive value and if downstream, enter a negative value.In order to go straight across, what angle upstream should she have headed?
Given
v = 2.90 mi/h
x = 1.20 mi
vs = 2.0 mi/h
θ = 25°
Procedure
Part a)
Velocity of the boat with respect to water stream is given as
[tex]\begin{gathered} v_b=v-v_s \\ v_b=(2.90\cdot\sin 25-2.00)\hat{j}+2.90\cdot\cos 25\hat{i} \\ v_b=-0.77\hat{j}+2.62\hat{i} \end{gathered}[/tex]magnitude of the speed is given as
[tex]\begin{gathered} v_b=\sqrt[]{0.77^2+2.62^2} \\ v_b=2.73mi/h \end{gathered}[/tex]Part b)
Time to cross the river is given as:
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{1.2mi}{2.90\cdot\cos25} \\ t=0.46h \end{gathered}[/tex]Part c)
Distance moved by boat in downstream is given as
[tex]\begin{gathered} x=v_yt \\ x=-0.77\cdot0.46 \\ x=-0.354mi \end{gathered}[/tex]Part d)
In order to go straight, we must net speed along the stream must be zero
so we will have
[tex]\begin{gathered} v\sin \theta=v_s \\ 2.90\sin \theta=2.00_{} \\ \sin \theta=\frac{2.00}{2.90} \\ \theta=43.60^{\circ} \end{gathered}[/tex]8) If the volume of the liquid in graduated cylinder B is 90 mL, then whatis the volume of the rock?AYour answer8060B100180
Answer:
20 mL
Explanation:
The volume of the rock is equal to the difference of volume of A and B. So, it is equal to
90 mL - 70 mL = 20 mL
Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.
Therefore, the volume of the rock is 20 mL
A cat chases a mouse across a 0.66 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor taylor (jdt3899) – Homework 3, 2d motion 22-23 – tejeda – (LermaHPHY1 1) 3 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table?
The cat’s speed when it slid off the table will be 6.552 m/s
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
a = -g = 9.8 m[tex]/s^{2}[/tex]
using equation of motion
x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]
since , a (horizontal) = 0
x = u(horizontal )*t
u = x / t equation 1
similarly
y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]
u(vertical) = 0
t = [tex]\sqrt{2y / a}[/tex] equation 2
substituting the value of equation 2 in equation 1
u = x / [tex]\sqrt{2y / a}[/tex]
= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]
= 6.552 m/s
The cat’s speed when it slid off the table will be 6.552 m/s
To learn more about kinematics here:
https://brainly.com/question/27126557
#SPJ1
Hunter pushed a couch across the room. He did 800 J of work in 20 seconds.The couch weighed 500 N. How much power did he have?A. 16,000 WB. 1.6WC. 800 WD. 40 W
Power = 40 W
Option D
Explanation:The workdone by the Hunter = 800 Joules
The weight of the couch = 500 N
Time, t = 20 seconds
Power = Workdone/Time
Power = 800/20
Power = 40 W
Ms. Terry travels to another planet where she drops a hamburger freely from rest. On this particular planet, g=5 m/s2. How far has the hamburger fallen after 4 s?
Answer:
40 m
Explanation:
d = 1/2 a t^2
= 1/2 * 5 m/s^2 * (4 s)^2 = 1/2 * 5 * 16 = 40 m
The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]