Answer:
Because there was fewer fossils
Explanation:
Answer:
Actually the answer is "The stratum contains iridium.".
Explanation:
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
A 4.0-kilogram ball held 1.5 meters above the floor has ________ joules of potential energy
Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.
What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?
Answer:
(a) the unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart is 317 m/s²
(c) the average acceleration of the lighter cart is 539 m/s²
Explanation:
Given;
mass of the first cart, m₁ = 0.66 kg
initial speed of the first cart, u₁ = 1.85 m/s
let the mass of the cart with unknown inertia be m₂
initial velocity of the second cart, u₂ = 2.17 m/s to the left
velocity of the first cart after collision, v₁ = 1.32 m/s to the left
velocity of the second cart after collision, v₂ = 3.22 m/s
time of collision, t = 0.010 s
(a) What is the unknown inertia?
Apply the principle of conservation of linear momentum, to determine the unknown inertia.
let leftward direction be negative direction
let rightward direction be positive direction
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)
1.221 - 2.17m₂ = -0.8712 + 3.22m₂
1.221 + 0.8712 = 3.22m₂ + 2.17m₂
2.0922 = 5.39m₂
m₂ = 2.0922 / 5.39
m₂ = 0.388 kg
The unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart
the heavier cart has a mass of 0.66 kg
[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]
(c) the average acceleration of the lighter cart;
the lighter cart has a mass of 0.388 kg
[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable
Please select the best answer from the choices provided
Answer:
B
Explanation:
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?
Answer:
1.9kHz
Explanation:
Given data
wavelength [tex]\lambda= 0.32m[/tex]
velocity [tex]v= 622 m/s[/tex]
We know that
[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]
substitute
[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]
Hence the frequency is 1.9kHz
Answer:
971.2
Explanation:
It was right on acellus :)
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy
Answer:
isnt heat transfer
Explanation:
sorry if im wrong
How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)
Answer:
Force = 2240 Newton.
Explanation:
Given the following data;
Mass A= 65kg
Mass B = 215kg
Acceleration = 8m/s²
To find the force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
First of all, we would have to find the total mass.
Total mass = Mass A + Mass B
Total mass = 65 + 215
Total mass = 280kg
Substituting into the equation, we have
[tex] Force = 280 * 8 [/tex]
Force = 2240 Newton.
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B with silk. Nylon is used to rub and charge a third piece of amber. Piece A and B are both repelled by the third piece of amber. This means:____.
Answer:
ieces A and B must also have the same type of charges
Explanation:
In electrostatics, charges of the same sign repel and charges of different signs attract.
If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.
Also part A and C repel each other, therefore they have the same type of charge.
If we use the transitive property of mathematics, pieces A and B must also have the same type of charges
A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
Answer:
Explanation:
equation of wave is given by the following equation
y = (2.6 cm) sin[1.8 - (5.8 s-1)t].
Comparing it with standard form of wave
y = A sin ( ωt - kx )
we get
ω = 5.8
2πn = 5.8
n = .92 per second
kx = 1.8
k x 6 = 1.8
k = 0.3
[tex]\frac{2\pi}{\lambda}[/tex] = 0.3
λ = 20.9 cm
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.
Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Answer:
Explanation:
kinetic energy = 14.1 MJ = 14.1 x 10⁶ J
Let radius of flywheel be r .
volume of flywheel = π r² x t where t is thickness
= 3.14 x r² x .113 m³
= .04 r² m³
mass = volume x density
= .04 r² x 7800 = 312.73 r²kg
moment of inertia I = 1 / 2 mass x radius²
= .5 x 312.73 r² x r²
= 156.37 r⁴ kg m²
angular velocity ω = 2π x 93/60
= 9.734 rad /s
kinetic energy = 1/2 Iω² where ω is angular velocity
= .5 x 156.37 r⁴ x 9.734²
= 7408.08 r⁴
Given
7408.08 r⁴ = 14.1 x 10⁶
r⁴ = .19 x 10⁴
r = .66 x 10
= 6.60 m .
Diameter = 13.2 m
b )
centripetal acceleration of a point on its rim = ω² r
= 9.734² x 6.6
= 625.35 m /s²
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Answer:
average speed is 1 m/s and average velocity is 0 m/s.
Explanation:
Given that :
Length of round trip = 50 m
Time taken = 100 seconds
The average speed :
Total distance / total time taken
Length of complete round trip :
(50 + 50) m, total. Distance = 100 m
100 / 100 = 1m/s
The average velocity :
Total Displacement / total time taken
Total Displacement of round trip = end point - start point = 0
0 / 100 = 0
Average speed is 1 m/s and average velocity is 0 m/s.
The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.
The speed is obtained from;
Speed = Distance/time = 2(50 m)/100 s = 1 m/s.
The velocity is 0 m/s since it is complete round-trip lap.
Learn more about speed: https://brainly.com/question/7359669
Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.
1. (6x + 8)(5x - 8)
a. 30x2 + 49x + 20
2. (5x + 6(5x - 5)
b. 24x3 + 8x2 + 6x + 4
3. (6x + 3)(6x - 4)
c. 25x2 + 5x - 30
4. (6x + 5)(5x + 5)
d. 30x2 - 8x - 64
e. 36x2 - 6x - 1
5. (4x + 2) (6x2 - x + 2)
Answer:
form 1 question??????????
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
How could a change in straight line motion due to unbalanced forces be predicted from an understanding of inertia?
Answer:
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
Explanation:
The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.
Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
