Answer:
Explanation:
The total distance travelled by ball before first bounce
= 160 + 160 = 320 ft
Distance travelled between first bounce and second bounce
= 80 + 80 = 160 ft
Distance travelled between second bounce and third bounce
= 40 + 40 = 80 ft
Distance travelled between third bounce and fourth bounce
= 20 + 20 = 40 ft .
Distance travelled in fourth bounce = 10 ft
Total distance travelled = 320 + 160 + 80 + 40 + 10
= 610 ft .
What is the function
of second plate in
parallel plate capacitor?
I need this done by tonight!! Can anyone help me please? Answer these 4 questions
Answer:
1. 14 g of chocolate mixture.
2. 24 fl oz of chocolate milk
3. 10 cups of chocolate milk.
4. 12½ cups.
Explanation:
From the question given above, the following data were obtained:
1 TBSP = 7 g
1 Cup = 8 fl oz
2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.
1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.
From the question given above,
1 Cup required 2 Table spoons (TBSP)
But
1 TBSP = 7 g
Therefore,
2 TBSP = 2 × 7 = 14 g
Thus, 1 Cup required 14 g of chocolate mixture.
2. Determination of the number fl oz of chocolate milk in 3 cups
1 Cup = 8 fl oz
Therefore,
3 Cups = 3 × 8
3 Cups = 24 fl oz
Thus, 24 fl oz of chocolate milk are in 3 cups.
3. Determination of the number of cups of chocolate milk produce from 20 TBSP.
2 TBSP is required to produce 1 cup.
Therefore,
20 TBSP will produce = 20/2 = 10 Cups.
Thus, 10 cups of chocolate milk produce from 20 TBSP.
4. Determination of the number of cups obtained from 100 fl oz chocolate milk.
8 fl oz is required to produce 1 cup.
Therefore,
100 fl oz will produce = 100 / 8 = 12½ cups.
Thus, 12½ cups is obtained from 100 fl oz chocolate milk.
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and released, calculate;the spring constant,the angular velocity, amplitude of oscillation, maximum velocity, maximum acceleration of the mass,period, frequency
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substance by 1 °C. Specific heat capacity can be calculated using the following equation:
q = mc deltaT
In the equation q represents the amount of heat energy gained or lost in joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and AT is the temperature change of the substance in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
Solve: When you mix two substances, the heat gained by one substance is equal to the heat lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with 1,000 g of water. The water changes temperature by 2 °C and the aluminum changes temperature by -74.95 °C.
A. Water has a known specific heat capacity of 4.184 J/g °C. Use the specific heat equation to find out how much heat energy the water gained (q).
B. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. Aluminum's accepted specific heat value is 0.900 J/g °C. Use this value to check your work.
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
The heat gained by the water can be obtained solving the following equation:[tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]
where cw = specific heat of water = 4.184 J/gºCm= mass of water = 1,000 gΔT = 2ºC Replacing these values in (1) we get:[tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]
B)
Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:[tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]
⇒ [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]
[tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]
which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.
help me help me help me
Which has a greater momentum and greater kinetic energy-a truck with
a mass of 3530 kg moving at a speed of 21 m/s or a car with a mass of
1620 kg moving at a speed of 54 m/s? Answers below are given in the
form: greater momentum, greater kinetic energy.
Answer:
the car has greater momentum.
the car has greater kinetic energy.
Explanation:
FOR MOMENTUM:
Momentum is given as the product of mass and velocity of an object:
[tex]P = mv[/tex]
where,
P = momentum
m = mass
v = velocity
For Truck:
[tex]P_{truck} = (3530\ kg)(21\ m/s)\\P_{truck} = 74130 Ns[/tex]
For Car:
[tex]P_{car} = (1620\ kg)(54\ m/s)\\P_{car} = 87480\ Ns[/tex]
Therefore, car has greater momentum.
FOR KINETIC ENERGY:
Kinetic Energy is given as:
[tex]K.E = \frac{1}{2} mv^{2}[/tex]
where,
K.E = Kinetic Energy
m = mass
v = velocity
For Truck:
[tex]K.E_{truck} = \frac{1}{2} (3530\ kg)(21\ m/s)^{2}\\K.E_{truck} = 778365\ J = 778.36\ KJ[/tex]
For Car:
[tex]K.E_{truck} = \frac{1}{2} (1620\ kg)(54\ m/s)^{2}\\K.E_{truck} = 2361960\ J = 2361.96\ KJ[/tex]
Therefore, car has greater kinetic energy.
Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
To learn more about soundwave, here:
https://brainly.com/question/31851162
#SPJ4
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
A bottle rocket is fired off and has an acceleration of 14.5 m/s2 for the 2.25s until it burns out. If it starts at rest, what distance does it cover?
Answer:
S = 16.3125m
Explanation:
Given the following data;
Acceleration, a = 14.5m/s²
Time, t = 2.25secs
Since the bottle rocket starts from rest, its initial velocity is 0m/s.
To find the distance S, we would use the second equation of motion.
S = ut + ½at²
Substituting into the equation, we have
S = 0(2.25) + ½*14.5*2.25
S = 0 + 7.25*2.25
S = 16.3125m
Therefore, the bottle rocket covered a distance of 16.3125 meters.
What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
describe measurement in our daily life
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
Learn more about kinetic energy here: https://brainly.com/question/1932411
Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
Answer:
Analysis
Explanation:
Because you must Analysis each and every cold too find out which virus caused this.
It’s weird because Interpretation and Analysis have the meaning of examination
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
A swift moving hawk is moving due west with a speed of 30 m/s; 5.0 s later it is moving due north with a speed of 20 m/s.
(A) What are the magnitude and direction of Δvav vector during this 5 s interval? and
(B) What are the magnitude and direction of vector aav during this 5 s interval?
Answer:
See explanation
Explanation:
Now we have;
vi = (-30, 0) m/s
vf = (0, 20) m/s
Δvav= vf - vi= (30, 20) m/s
magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s
Direction = tan-1(20/30) = 33.69°
For aav
aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2
magnitude of aav = √6^2 + 4^2 = 7.2 m/s^2
direction of aav = tan-1(4/6) = 33.69°
What relationship must exist between an applied force and the velocity of a moving object if uniform circular motion is to result?
Answer:
See explanation
Explanation:
Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.
The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.
So we can write;
F = mv^2/r
If a car is traveling at an average speed of 20 m/s, how long will it take to travel 500 meters?
A. 0.04 seconds
B. 25 seconds
C. 520 seconds
D. 10,000 seconds
Answer:
B. 25 seconds
Explanation:
500÷20=25
If an atom of oxygen has an atomic number of eight that means...…
E. there are 8 protons
F. there are 8 neutrons
G. it weighs 8 amu
H. it is in group 8
explain resolution of Force
Answer:
it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.
Explanation:
How do I solve this problem please help
Answer:
som,eythiung did cant wait
Explanation:
ya so i texted maria and she said i con c()me over so letyeyey yeah betty letty
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