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
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Before they were decommissioned, the NASA space shuttles required two solid rocket boosters (SRBs) to launch the shuttle from Earth’s surface. Both SRBs produced 1.7 x10^7 N at liftoff. The combined mass of a shuttle and rocket boosters was about 1.5 x 10^ 6 kga) Calculate the net acceleration of a space shuttle and rockets at the time of liftoff. (b) Calculate the speed of the shuttle and rockets after 10.0 s.
Given:
The force on the booster is
[tex]F=\text{ 1.7}\times10^7\text{ N}[/tex]The mass is
[tex]m=\text{ 1.5}\times10^6\text{ kg}[/tex]Required:
(a) The net acceleration
(b) Speed of the shuttle and rockets after time t = 10 s
Explanation:
(a) The net acceleration can be calculated as
[tex]\begin{gathered} a=\frac{F}{m} \\ =\frac{1.7\times10^7}{1.5\times10^6} \\ =11.33\text{ m/s}^2 \end{gathered}[/tex](b)
The initial speed of the rocket and shuttle will be zero.
The speed of the rocket and shuttle after time t = 10 s will be
[tex]\begin{gathered} v=0\text{ m/s + 11.33}\times10 \\ =\text{ 113.3 m/s} \end{gathered}[/tex]Final Answer:
(a) The net acceleration is 11.33 m/s^2.
(b) The speed of the rocket and shuttle is 113.3 m/s
Use the acceleration vs time graph to answer this question. The graph shows the motion with an initial velocity of -4 m/s. Each tick mark on the x-axis represents 1 second. Calculate the velocity at t = 8 seconds.
Answer:
4 m/s
Explanation:
To find the velocity at t = 8 seconds, we will use the following equation:
[tex]v_f=v_i+at[/tex]Where vf is the final velocity, vi is the initial velocity, a is the acceleration and t is the time.
From t = 0 seconds to t = 3 seconds, we have an acceleration of 6 m/s², so we can calculate the velocity at t = 3 seconds as:
[tex]\begin{gathered} v_f=-4m/s+6m/s^2(3\text{ s)} \\ v_f=-4\text{ m/s + 18 m/s} \\ v_f=14\text{ m/s} \end{gathered}[/tex]Now, from t = 3 seconds to t = 8 seconds, the acceleration is equal to -2 m/s². So we need to use the same equation but this time, the initial velocity will be 14 m/s and the time will be 5 seconds because t = 8 s - 3 s = 5s. Then, we get:
[tex]\begin{gathered} v_f=14m/s-2m/s^2(5s) \\ v_f=14\text{ m/s - 10 m/s} \\ v_f=4\text{ m/s} \end{gathered}[/tex]Therefore, the velocity at t = 8 seconds is 4 m/s
The potential energy of a system can be changed by varying the position of objects in the system. At which point do the coaster cars have the most potential energy?
To find:
At which point the potential energy of the coaster car is the highest?
Explanation:
The potential energy of an object is the energy possessed by the object due to the position of the object. The gravitational potential energy of the object is directly proportional to the height of the object.
Thus an object will have the highest potential energy when it is at the highest point.
Final answer:
Thus the coaster cars will have the highest potential energy when it is at the highest point on the roller coaster.
Therefore, the coaster cars will have the most potential energy at point A.
A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a speedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
[tex]\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}[/tex]Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
[tex]\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}[/tex]Therefore, the average force exerted on the car by the wall was 44200 N
A girl walks 600 m north and then 800 m east. What is the displacement from her starting point?
Displacement from her starting point is 529m
What is Displacement?
"Displacement" describes a change in an object's location. It is a vector quantity since it has a magnitude and a direction. It looks like an arrow that leads from the beginning to the end.
Given : A girl walks 600m north
She then walks 800m east
To Find : Displacement from her starting point
Solution: North
West East
South
North=600m
East =800m
We use Pythagoras Theorem
Displacement to be covered = [tex]\sqrt{800^{2} }[/tex]- [tex]\sqrt{600^{2} }[/tex]
529m
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carts, bricks, and bands
7. Which of the following conclusions are supported by the data in Table 2?
a. Adding bricks to a cart has no affect upon the cart's acceleration.
b. Increasing the mass of an object causes a decrease in its acceleration.
c. An increase in the number of rubber bands causes an increase in the acceleration.
d. The more mass that an object has, the more acceleration that it will acquire when pushed.
The conclusions that are specifically supported by the data in Table 2 Increasing the mass of an object causes a decrease in its acceleration. That is option B.
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 5 ----> no bricks = 0.99 m/s²
Trial 6 ----> cart with one brick = 0.50 m/s²
Trial 7 ----> cart with two bricks = 0.32 m/s²
Trial 8 -----> cart with three bricks = 0.25 m/s²
From the information above, progressive increase the the quantity and mass of the bricks lead to a decrease in the acceleration of the cart with a constant force from only 2 bands.
This occurred because the mass is inversely proportional to acceleration.
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The cable is drawn into the motor with an acceleration of 3 m/s2 .
Determine the time needed for the load at B to attain a speed of 10 m/s, starting from rest.
The time needed for the load at B to attain a speed of 10 m/s, starting from rest is 3.33 s
How to determine the timeAcceleraion is defined according to the following formula:
a = (v – u) / t
a is the acceleration v is the final velocity u is the initial velocity t is the timeWith the above formula, we can determine the time. Details below.
The following data were obtained from the question
Acceleration (a) = 3 m/s² Initial velocity (u) = 0 m/sFinal velocity (v) = 10 m/sTime (t) =?a = (v – u) / t
3 = (10 – 0) / t
3 = 10 / t
Cross multiply
3 × t = 10
Divide both sides by 3
t = 10 / 3
t = 3.33 s
Thus, we can say that the time required to attain the velocity of 10 m/s is 3.33 s
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Missing part:
See attached photo
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
A maximum of about 229 repetitions of something like the exercise can be performed by that of the athlete utilizing the energy provided by each of the biscuits.
The proportion of input to produced energy can be used to define energy consumption.
A cookie, therefore, therefore has 54.0 kcal of calories. The 54.0 kcal throughout this croissant is used as power input by the athlete.
Efficiency = output energy / input energy
It can be written as:
Output energy = efficiency × input energy
Puting the values of efficiency and input energy.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
The weightlifting exercise can be done n times for the output energy. This outgoing energy comes from mgh in the shape of potential energy. So,
Energy per repetition = [tex]mgh[/tex]
Put the values of m, g and h in above equation.
Energy per repetition = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J
Energy per repetition = 0.059 kcal.
So,
amount of repetitions = sum of output energy / energy per repetition
amount of repetitions = 13.5 kcal / 0.059 kcal = 229 repetitions.
Therefore, amount of repetition can be be 229.
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What is the car's velocity between 11h and 15h?
Equation:
df-di/ t f-ti
The velocity of the car between time 11 hr and 15 hr which is shown in the graph would be 4 m/s².
The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
We are given that,
Displacement of the car = Δx = (20km) - (4km) = 16 km
Time interval of the car = Δt = (15h)- (11h) = 4hours
v = dx/dt
dx = v dt
∫dx = ∫v dt
Δx = v Δt
v = Δx/Δt
Therefore, for get the value of velocity between the given time interval , putting the value in in above equation,
v = 16km/4hours
v = 4 km/hours
Thus, The velocity of the car between time 11 hr and 15 hr will be given as 4km/hours.
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The virtual image as seen in a plane mirror is reversed both left-to-right and top-to-bottom. Is this true or false?
ANSWER:
false
STEP-BY-STEP EXPLANATION:
The virtual images in a plane mirror have a left-right investment. But not top-to-bottom, which means that what the statement says is false.
the answer to this question
The car and the delivery truck both start from rest and accelerate at the same rate. So, the final speed of the car as compared to the truck is four times as much. So, option D is correct.
What is acceleration?Acceleration is the term used in mechanics to describe how quickly an object's velocity changes in relation to time. The magnitude of accelerations as a vector. An object will accelerate in the direction that it is being pulled in by the net force.
Acceleration is a vector quantity since it has a magnitude and a direction. Velocity is another aspect of vessel quantities. The change in the vector of velocity over a time interval divided by the time interval is the definition of acceleration.
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Which of the following types of radiation consists of a high energy electron?A. GammaB. BetaC. DeltaD. Alpha
Explanation:
a) Gamma rays are high-energy photons and are the most energetic part of the electromagnetic spectrum.
b) Beta particles are high-energy, high-speed electrons emitted by certain radioactive nuclei.
c) There is no radiation called delta radiation.
d) Alpha particles are particles that are composed of two protons and two neutrons.
Final answer:
Thus, the correct option is (B) Beta
What is the density of a 45.87 g golf ball with a diameter of 4.287 cm?
We are asked to determine the density of a gulf ball given its mass and volume. To do that, we will use the formula for density:
[tex]D=\frac{m}{V}[/tex]Where:
[tex]\begin{gathered} D=\text{ density} \\ m=\text{ mass} \\ V=\text{ volume} \end{gathered}[/tex]To determine the volume we will use the fact that the gulf ball can be approximated to a sphere and the volume of a sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]Where:
[tex]r=\text{ radius}[/tex]We are given the diameter. We know that the diameter is twice the radius, therefore:
[tex]r=\frac{D}{2}[/tex]Substituting the value of the diameter we get:
[tex]r=\frac{4.287\operatorname{cm}}{2}[/tex]Solving the operations:
[tex]r=2.144\operatorname{cm}[/tex]Now, we substitute the value of the radius in the formula of the volume:
[tex]V=\frac{4}{3}\pi(2.144\operatorname{cm})^3[/tex]Solving the operation we get:
[tex]V=41.282\operatorname{cm}^3[/tex]Now, we substitute the value of the volume and the mass in the formula for density:
[tex]D=\frac{45.87g}{41.282\operatorname{cm}^3}[/tex]Solving the operation:
[tex]D=1.11\frac{g}{\operatorname{cm}^3}[/tex]Therefore, the density of the ball is 1.11 g/cm^3.
help, Asap!!!!!!!!!!!!!!!!!!
Answer:
0
Explanation:
all of the movement and opposite movement are the same
A train car with a mass of 10kg and speed of 10 m/s is traveling to the right. Another train car with a mass of 20kg is moving to the left at -40 m/s. After the collision, the 10 kg train car is now moving at -20 m/s and we need to find the Velocity of the 20 kg train car.
When two particles collide and the masses of the particles are given, as well as the initial and final velocity of one particle and one of the velocities of the second particle, then the remaining velocity of the second particle is given by the expression:
[tex]v_2^{\prime}=\frac{m_1v_1+m_2v_2-m_1v_1}{m_2}[/tex]Which can be deduced from the Law of Conservation of Linear Momentum.
In the given problem, the initial and final velocities of the train car with mass 10kg are given, as well as the initial velocity of the 20kg car:
[tex]\begin{gathered} m_1=10kg \\ v_1=10\frac{m}{s} \\ v_1^{\prime}=-20\frac{m}{s} \\ \\ m_2=20kg \\ v_2=-40\frac{m}{s} \\ v_2^{\prime}=\text{ unknown} \end{gathered}[/tex]Replace those values into the given equation to find v₂':
[tex]\begin{gathered} v_2^{\prime}=\frac{(10kg)(10\frac{m}{s})+(20kg)(-40\frac{m}{s})-(10kg)(-20\frac{m}{s})}{20kg} \\ \\ \Rightarrow v_2^{\prime}=-25\frac{m}{s} \end{gathered}[/tex]Therefore, the velocity of the 20kg train car after the collision, is: -25 m/s.
What is the normal force of a 0.0037 kg tennis ball rolling at a constant speed of 3 m/s
across a desk?
The normal force of the rolling ball is 0.037 N
Mass of tennis ball= 0.0037 kg
Constant speed= 3m/s
we need to apply the concept of laws of motion
Since the ball is rolling at a constant speed, it is an example of uniform motion.
So,
Normal force=weight of the body
= 0.0037 x 10 ( acceleration of gravity= 10 m/s²)
Normal force= 0.037 N
Therefore the normal force on the ball is 0.037 N
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An object has an excess charge of −1.6 × 10−17 C. How many excess electrons does it have?
Given:
The charge on the object is
[tex]q=-1.6\times10^{-17}\text{ C}[/tex]Required: Number of electrons
Explanation:
The number of electrons can be calculated using the quantization of charge
[tex]q\text{ = ne}[/tex]Here, n is the number of electrons
e is the charge on the electron whose value is
[tex]e\text{ = -1.6}\times10^{-19}\text{ C}[/tex]
On substituting the values, the number of electrons will be
[tex]\begin{gathered} n=\frac{q}{e} \\ =\frac{-1.6\times10^{-17}}{-1.6\times10^{-19}} \\ =100 \end{gathered}[/tex]Final Answer: The object has an excess of 100 electrons.
A ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22.0%,
If a ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22°, and the refractive index of the liquid would be 1.334.
What is refraction?It is the phenomenon of bending of light when it travels from one medium to another medium. The bending towards or away from the normal depends upon the medium of travel as well as the refractive index of the material.
By using Snell's law,
Refractive index of the liquid = sin(i) /sin(r)
=sin(30) /sin(22)
= 1.334
Thus, the refractive index of the liquid would be 1.334.
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QUESTION 22As the rocket moves from position "b" to posisition "c", its speed is:constant.O continuously increasing.O continuously decreasing.increasing for a while and constant thereafter.constant for a while and decreasing thereafter.
The rocket movement due to the direction between a and b, and turning on the engine, movement is described by E image
Is continuously increasing, the force is constant and the acceleration too, which means that the speed continues increasing
Vector A= 30 m/s towards East and vector B= 80 m/s towards south. Find A- B [Perform the subtraction of the vector].
ANSWER
[tex]\begin{equation*} 85.44\text{ }m\/s \end{equation*}[/tex]EXPLANATION
First, let us make a sketch of the two vectors:
The vector A - B is represented by line BA in the figure above.
To evaluate A - B, we will apply the Pythagoras theorem:
[tex]\begin{gathered} A-B=BA=\sqrt{(30)^2+(80)^2} \\ A-B=BA=\sqrt{900+6400} \\ A-B=BA=\sqrt{7300} \\ A-B=BA=85.44\text{ }m\/s \end{gathered}[/tex]That is the answer.
A 4000-kg truck traveling with a velocity of 20 m/s due south collides head-on with a 1320- kg car traveling with a velocity of 10 m/s due north. The two vehicles stick together after the collision.A. What are the magnitude and the direction of the momentum of each vehicles Prior to the collision?B. What are the magnitude and the direction of the velocity of both vehicles after their collide?
Before we begin, we will establish that the north direction is the positive direction which means that the south direction is negative.
A.
The momentum of an object is given by:
[tex]p=mv[/tex]For the truck we know that the velocity is -20 m/s (since it is traveling south) and its mass is 4000 kg, then its momentum is:
[tex]p=(4000)(-20)=-80000[/tex]Therefore, the momentum of the truck is -80000 kg m/s; this means that its magnitude is 80000 kg m/s and its direction is south.
For the car we know that the velocity is 10 m/s and its mass is 1320 kg, then its momentum is:
[tex]p=(1320)(10)=13200[/tex]Therefore, the momentum of the car is 13200 kg m/s which means that its magnitude is 13200 kg m/s and its direction is north.
B.
In a collision the momentum is conserved, that is, the total initial and final momentum is equal, that is:
[tex]p_i=p_f[/tex]In this case, we know that the vehicles stick together after they collide, then we have:
[tex]m_tv_t+m_cv_c=(m_t+m_c)u[/tex]where u is the velocity of the vehicles after they collide. Plugging the values we know, we have that:
[tex]\begin{gathered} -80000+13200=(4000+1320)u \\ u=\frac{-80000+13200}{4000+1320} \\ u=-12.56 \end{gathered}[/tex]Therefore, the final velocity of the system is -12.56 m/s which means that the magnitude of the velocity is 12.56 m/s and its direction is south.
An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?
Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
why is the hubble space telescope in space instead of on the ground?
Answer:EARTH'S ATMOSPHERE ALTERS AND BLOCKS THE LIGHT THAT COMES FROM SPANCE.
Explanation:
Hubble orbits above Earth's atmosphere, which gives it a better view of the universe than telescopes have at ground level.
The displacement (in meters) of a particle moving in a straight line is given by S= t^2 - 7t + 17 ii)iii)iv)
Given displacement of a particle moving in a straight line,
[tex]S=t^2-7t+17[/tex](i) Calculate the average velocity in the interval [3,4]
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4)-S(3)}{4-3} \\ \text{Average velocity = }\frac{\lbrack(4)^2-7\times4+17\rbrack-\lbrack(3)^2-7\times3+17\rbrack}{4-3} \\ \text{Average velocity =}\frac{5-5}{1} \\ \text{Average velocity = 0 m/s} \end{gathered}[/tex](iii) Calculate the average velocity in the time interval [4,5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(5)-S(4)}{5-4} \\ \text{Average velocity = }\frac{\lbrack(5)^2-7\times5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{5-4} \\ \text{Average velocity = }\frac{7-5}{1} \\ \text{Average velocity = }2 \end{gathered}[/tex](iv) Calculate the average velocity in the time interval [4,4.5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4.5)-S(4)}{4.5-4} \\ \text{Average velocity = }\frac{\lbrack(4.5)^2-7\times4.5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{4.5-4} \\ \text{Average velocity = }\frac{5.75-5}{0.5} \\ \text{Average velocity = }1.5 \end{gathered}[/tex]A man can crack a nut by applying a force of 100 N a lever of length 0.5 m. What should be the length of the lever if he wants to use a force of 75 N to crack the nut?
Answer:
So, to apply the force of 75N length of nut cracker should be 66.6 cm
Answer:
Explanation:
Given:
F₁ = 100 N
L₁ = 0.5 m
F₂ = 75 N
__________
L₂ - ?
According to the rule of moments:
M₁ = M₂
F₁·L₁ = F₂·L₂
The length of the lever:
L₂ = F₁·L₁ / F₂
L₂ = 100·0.5 / 75 ≈ 0.67 m
Which is negatively charged?A. protonB. nucleusC. electronD. neutron
Protons, electrons an neutrons are the particles that make up atoms.
Protons have a positive electric charge, electrons have a negative electric charge an neutrons are electrically neutral.
The nucleus of an atom is made of protons and neutrons, so, its electric charge is positive an proportional to th
a piece of steel expands 10 cm when it is heated from 20 to 40 deg C. How much would it expand if it was heated from 20 to 60 deg.C?
ANSWER
20 cm
EXPLANATION
Given:
• The change in length of a piece of steel, ΔL = 10 cm, when its temperature change is ΔT = 20°C
Find:
• The change in length of the same piece of steel, ΔL, when its temperature is changed ΔT = 40°C
The change in length of a solid material is,
[tex]\Delta L=\alpha\cdot L_o\cdot\Delta T[/tex]Where L₀ is the original length, α is the linear expansion coefficient and ΔT is the change in temperature.
In this case, we know how much is the piece of steel expanded when the temperature change is 20°C, so we can find the product αL₀,
[tex]\alpha L_o=\frac{\Delta L}{\Delta T}[/tex]Replace the known values and solve,
[tex]\alpha L_o=\frac{10\text{ }cm}{20\degree C}=0.5cm/\degree C[/tex]If the temperature change now is 40°C,
[tex]\Delta L=\alpha L_o\Delta T=0.5cm/\degree C\cdot40\degree C=20cm[/tex]Hence, when the piece of steel is heated from 20°C to 60°C it will expand 20 centimeters.
A 3 kg ball is dropped from a height of 100 m above the surface of Planet Z. If the ball reaches a velocity of 45 m/s in 7 s, what is the ball’s weight on Planet Z? What is the gravitational field strength on Planet Z?
We are given the following information
Mass of ball = 3 kg
Height = 100 m
Final velocity = 45 m/s
Time = 7 s
Recall from the equations of motion
[tex]s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2[/tex]Where u is the initial velocity of the ball that is zero.
[tex]\begin{gathered} s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2 \\ 100=0\cdot7+\frac{1}{2}\cdot g\cdot7^2 \\ 100=\frac{1}{2}\cdot g\cdot49 \\ g=\frac{2\cdot100}{49} \\ g=4.08\; \frac{m}{s^2} \end{gathered}[/tex]So, the gravitational acceleration of the planet Z is 4.08 m/s^2
The weight o the ball is given by
[tex]\begin{gathered} W=m\cdot g \\ W=3\cdot4.08 \\ W=12.24\; N \end{gathered}[/tex]Therefore, the weight of the ball is 12.24 N
What is the average velocity of a car that travels 48 km north in 2.0 h?
The average velocity of the car would be 24 kilometers per hour in the north direction if the car travels 48 kilometers in the north for 2 hours.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem, we have to find the average velocity of the car if the car travels 48 kilometers in the north for 2 hours.
The average velocity of the car = 48 kilometers / 2 hours
= 24 kilometers per hour
Thus, the average velocity of the car would be 24 kilometers per hour
Learn more about Velocity here, refer to the link ;
brainly.com/question/18084516
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i’m still really confused on how to actually calculate it
Question 6:
Given information:
Distance travelled by bus,
[tex]s=10100\text{ m}[/tex]Average velocity of the bus,
[tex]v=5.6\text{ m/s}[/tex]We need to find the time taken by bus to reach school. Let t be the time taken by bus to reach school. The velocity of the bus is given as,
[tex]v=\frac{s}{t}[/tex]The expression for the time is given as,
[tex]t=\frac{s}{v}[/tex]Substituting all known values,
[tex]\begin{gathered} t=\frac{10100\text{ m}}{5.6\text{ m/s}} \\ \approx1804\text{ s} \\ \approx30\text{ min 4 sec} \end{gathered}[/tex]Therefore, the bus required 30 min 4 sec to reach school.