Typing in the commands and instructions and pressing enter will put a new command in the command prompt.
What is a Command?This refers to a words or phrase which causes the computer to execute certain tasks or functions.
Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.
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PLEASE HELP!! 50 points. The answer is not 2.6 m/s.
Answer:
A
Explanation:
5.2mm/s
hope this helps
the road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60 km / h. Find the average round trip speed.
Answer:
Let distance between two cities is x km . So total distance travelled bus x+x=2x km and time taken was=(x/40)+(x/60)=(3x+2x)/120=5x/120=x/24 hrs . So avg speed is 2x/(x/24)=48 km per hr.
or
dt1=60km1h
dt2=30km1h
2dt1+t2=r¯
d=60t1
d=30t2
60t1=30t2
2t1=t2
2dt1+(2t1)=r¯
2(60t1)t1+2t1=r¯
2(60t1)=r¯(t1+2t1)
120=r¯+2r¯
r¯=1203
r¯=40km/h
Question 4 of 20
If the distance between two objects is doubled, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It increases to 2 times its original magnitude.
C. It increases to 4 times its original magnitude.
D. It decreases to 1/4 its original magnitude.
Answer:
D
Explanation:
F = G m1 m2 / r^2 now double r
F = G m1m1/ (2r)^2
F = 1/4 G m1m2/r^2 <===== this is 1/4 of the original
10 precautions for obtaining the refractive index of a triangular glass prism.
Answer:
Following precautions for obtaining the refractive index of a triangular glass prism:-
Explanation:
(1) All the faces of the prism should be neat and clean.
(2) Pins for holding the paper on the drawing board must be pinned perpendicular to the paper for better handling.
(3) While fixing the pins for checking the refractive index of the prism, make sure that the reflective images of the pins should be aligned to your eye to avoid any type of parallax error.
(4) Pins should not be removed or pinned again during the experiment.
(5) Avoid mishandling of the prism.
(6) Same edge of the prism should be taken as vertex for observations.
(7) For drawing the boundary of the prism, a sharp pencil should be used.
(8) Soft board and pointed pins should be used.
(9) The distance between the pins should be 5 cm or more.
(10) Make sure the glass slab or prism taken are polished and not broken.
Hope this helps!
Please mark me as brainlist.
A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
What is the specific heat capacity?The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
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Which of the following is not a use of nuclear chemistry in industry?
thickness gauges
tracers
leak detectors
stimulate plant growth
Stimulating plant growth is not a use of nuclear chemistry in industry. Thus, the correct option is D.
What is Nuclear Chemistry?Nuclear chemistry may be defined as the sub-branch of chemistry that involves the investigation of physical and chemical characteristics of elements that are affected by transformations in the configuration of the atomic nucleus.
Thickness gauges measure the thickness, density, and fill levels and are utilized in manufacturing to make sure that an entire product or a set of products are of identical thickness.
Tracers are utilized in tracing the actual composition of radioactive decay through the chemical compounds. While the leak detectors monitor the actual and constant flow of water through a given system precisely.
Therefore, the correct option for this question is D.
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Answer:
it is d
Explanation:
p ⇒ q, r ⇒ s, p ∨r q ∧ s is invalid
Answer:
My explanation and answer are down here↓:
Explanation:
p→q is false only when p is true and q is false.
∴p→(q∨r) is false when p is true and (q∨r) is false, and
q∨r false when both q,r are false.
Hence T,F,F.
A mass of (200 of hot water at (75, 0 degrees * C) is mixed with cold water of mass M at (5, 0 degrees * C) The final temperature of the mixture is (25, 0 degrees * C) What is the mass of the cold water (M)
Q in = Q out
mcΔt in = mcΔt out (c = equal)
mΔt in = mΔt out
200 x (75-25) = M x (25-5)
M = 500 g
∆T for hot water
75-2550°C∆T for cold water
25-520°CBoth are equal according law of conservation of energy
[tex]\\ \rm\Rrightarrow m_1c\delta T_0=m_2c\delta [/tex]
[tex]\\ \rm\Rrightarrow 200(50)=m_2(20)[/tex]
[tex]\\ \rm\Rrightarrow 10000=m_2(20)[/tex]
[tex]\\ \rm\Rrightarrow m_2=500g=0.5kg[/tex]
A mass of 2.0 kg of water is heated. The temperature increase of the water is 80 degrees Celsius. The specific heat capacity of water is 4200 J / kg degrees Celsius. Calculate the change in thermal energy when the water is heated. Use the equation. change in the thermal energy = mass x specific heat capacity x temperature change
Answer:
672 000 J or 672 kJ
Explanation:
change in the thermal energy = mass x specific heat capacity x temperature change
= 2 kg * 4200J/kg-C *80 C =672000 J
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?
(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]
where [tex]R[/tex] is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]
so that
[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]
Solve for [tex]v[/tex] :
[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]
(b) The net force equation in part (a) leads us to the relation
[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]
so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.
Identify the law, write the equation and calculate the answer to the problem below.
An incident light ray strikes water at an angle of 45 degrees. The index of refraction of air is 1.0003, and the index of refraction of water is 1.33. What is the angle of refraction rounded to the nearest whole number?
Find refractive index first
[tex]\\ \rm\Rrightarrow \mu=\dfrac{c}{v}[/tex]
[tex]\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}[/tex]
[tex]\\ \rm\Rrightarrow \mu =0.75[/tex]
Now
[tex]\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr[/tex]
[tex]\\ \rm\Rrightarrow sinr=0.94[/tex]
[tex]\\ \rm\Rrightarrow r=sin^{-1}(0.94)[/tex]
[tex]\\ \rm\Rrightarrow r=70^{\circ}[/tex]
What is the magnitude of the gravitational force of attraction to Jupiter exerts on IO
The gravity force between Jupiter and Io will be 6.343 × 10²² N.
What is Newton's law of gravitation?
Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.
Given data;
Mass of Jupiter,[tex]\rm m_j = 1.9 \times 10^{27} \ kg[/tex]
Mass of moon of Jupiter,[tex]\rm m_{i_0}= 8.9 \times 10^{22} \ kg[/tex]]
The gravitational constant is,[tex]\rm G = 6.67 \times 10^{-11 } \ m^3 kg^{-1} s^{-2}[/tex]
Distance between Jupiter and Io, R = 421,700 km = 4,217,00,000 m
The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.
The gravitational force is found as;
[tex]\rm F = G \frac{ m_J m_{I_0}}{R^2} \\\\\ F = (6.67\times 10^{-11}) \frac{( (1.9\times 10^{27})\times (8.9\times 10^{22} )} { (421700000)^2}\\\\ F_g = 6.343 \times 10^{22} \ N[/tex]
Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.
The complete question is
"Jupiter has a mass of 1.9 × 1027 kg, and its moon Io has a mass of 8.9 × 1022 kg. Their centers are separated by a distance of 421,700 km what is the force of gravity acting on Io? "
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Answer asap for brainlist
They use less wire, A
What is the mass of an object if it has 915J of energy when it is 6.50m above the ground?
A.851kg
B0.0696kg
C.583kg
D.14.4kg
Answer:
D
Explanation:
GPE = mgh
GPE / gh = m
m = GPE/gh
m = 915 / 6.5 x 9.8
m = 915 / 63.7
m = 14.4kg
At a given temperature, lighter gas molecules:
A) have an average speed less than heavier gas molecules.
B) have an average speed greater than heavier gas molecules.
C) ll move more slowly than heavier gas molecules.
At a given temperature, lighter gas molecules, have an average speed greater than heavier gas molecules.
What is temperature?Temperature is the measure of degree of hotness or coldness of a body. It is also the measure of average kinetic energy of a body.
As temperature increases, the average kinetic energy of a body increases.
K.E = ¹/₂mv²
v² = 2K.E/m
v = √( 2K.E/m)
where;
m is massv is velocityK.E is kinetic energyFrom the equation above, at a constant temperature, the speed of a gas molecule will decrease with increase in the mass of the molecules and vice versa.
Thus, at a given temperature, lighter gas molecules, have an average speed greater than heavier gas molecules.
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The three waves shown in the picture below are equal in wavelength, but are travelling at different speeds. Given this information, which wave has the highest frequency?
Select one:
a. 1
b. 2
c. 3
d. Can't tell from the information given.
Wave 3 will have the highest frequency. Option c is correct.
What is the frequency?
Frequency is defined as the number of repititions of a wave occurring waves in 1 second.
Frequency is given by the formula as,
[tex]\rm v = \lambda \times f[/tex]
If the wavelength of the wave is content the velocity of the wave is directly propotional to the frequency.
The velocity of wave 3 is maximum. So that the wave 3 will have the highest frequency.
Hence option c is correct.
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pleaseeeeeee I need HELP in 20 mins,Asap!!!
A 6.00 -kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22m/s^2 and shatters into three pieces, which all fly backward. The wall exerts a force of 2640N on the ball of for 0.1s. One piece of mass 2kg travels backward at a velocity of 10m/s and an angle of 32° above the horizontal. A second piece of mass 1kg travels at a velocity of 8m/s and an angle of 28° below the horizontal. What is the velocity of the third piece?
The velocity of the third piece is 124.02 m/s at 1.05⁰ below the horizontal.
Velocity of the third piece
The velocity of the third piece is calculated from the principle of conservation of linear momentum.
mu = m₁u₁ + m₂u₂ + m₃u₃
where;
m is mass of the clayu is velocity of the clayu₁ is velocity of first pieceu₂ is velocity of second pieceu₃ is velocity of third piecem₃ is mass of the third piece = 6 kg - (2 kg + 1 kg) = 3 kgMomentum in y - direction6(22)sin(0) = 2(10)sin32 - 1(8)sin(28) + 3u₃y
0 = 6.84 + 3u₃y
u₃y = -6.84/3
u₃y = -2.28 m/s
Change in momentumΔP = Pf - Pi = J
where;
Pf is final momentumPi is the initial momentumJ is impulse2640(0.1) = 2(10)cos32 + 1(8)cos(28) + 3u₃x - 6(22)
264 = -108 + 3u₃x
3u₃x = 372
u₃x = 372/3
u₃x = 124 m/s
Resultant velocityu₃ = √(124² + 2.28²)
u₃ = 124.02 m/s
Direction of the velocitytanθ = u₃y/u₃x
tanθ = 2.28/124
tanθ = 0.018
θ = 1.05⁰ (below the horizontal)
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an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of the object?
The height of the object will be -5.19 cm
A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.
Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall
So let,
v = Image distance from the mirror = -33.5 cm
u = object distance from the mirror (concave) = 24 cm
hi = Image height = 7.25 cm
h = height of the object = ?
Using below formula to find height of the object
-v/u = hi/h
Putting all value in the formula we get
-(-33.5)/(-24) = 7.25/h
h = -5.19 cm
Therefore the height of the object will be -5.19 cm
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What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the ground?
-441J
735J
-735J
441J
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
what type of forces that created at shaping machine
Answer:
a magnetic force
Explanation:
a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot
The force the rod exerts on the ball at the given angle is determined as 3.94 N.
Force exerted on the rod by the ball
The force exerted is calculated as follows;
F = ma
F = mv²/r
F = mω²r
where;
m is mass of the ballω is angular speed of the ballr is radius of the pathr = 2cosθ
Angular speed when the ball moves 15 degreesωf² = ωi² + 2αθ
where;
θ is angular displacement in radians, 15⁰ = 15 x π/180 radωf² = (1)² + 2(2)(15 x π/180)
ωf² = 2.04
ωf = √2.04
ωf = 1.428 rad/s
F = mω²(2cosθ)
F = (1)(1.428)²(2 x cos15)
F = 3.94 N
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You are standing in an open field near a train track as a train goes by at a constant speed blowing its horn in all directions (except down, into the ground) at a constant frequency. You are first able to hear it when it is approaching you 1 km away. When it is a quarter of a kilometer past you, by how many decibels has the train's sound level intensity shifted?
(options: 6, 12, 58, 105, 108, or 114)
Answer:
108 is the correct answer
Calculate the frequency of the wave shown below.
Frequency = Speed/Wavelength (λ)
Select one:
a. 4 hertz
b. 3 hertz
c. 2 hertz
d. 1 hertz
[tex] \color{skyblue}{ \underline{ \frak { \: option \: ( \: c \: ) = 2 \: hertz ✓}}}[/tex]
[tex] \: \: [/tex]
Given :
Wavelength ( λ ) = 2 m[tex] \: \: [/tex]
Speed = 4 m/s[tex] \: \: [/tex]
We, have to find frequency :
[tex] \: [/tex]
[tex] \large \tt \: Frequency = \frac{Speed}{Wavelength ( \: λ \: )} [/tex][tex] \: \: [/tex]
[tex]\large \tt \: Frequency = \frac{4}{2} [/tex][tex] \: \: [/tex]
[tex]\large \tt \: Frequency = \cancel \frac{4}{2} [/tex][tex] \: \: [/tex]
[tex] \pink{ \boxed{\large \tt \: Frequency =2 \: Hertz ✓}}[/tex][tex] \: \: [/tex]
Hope Helps!
You drive 8.50 km in a straight line in a direction 30° East of North.
(a) Find the distances you would have to drive straight East and then straight North to arrive at the same point. (This is equivalent to finding the components of the displacement along the East and North directions.)
km East
km North
(b) Show that you still arrive at the same point if the East and North legs are reversed in order.
(a.1) The component of displacement along the East is 7.36 km.
(a.2) The component of displacement along the North is 4.25 km.
(b) When the East and North legs are reversed in order, you will still travel 8.5 km.
Component of displacement along the East
dx = d cosθ
dx = 8.5 km x cos(30)
dx = 7.36 km
Component of displacement along the Northdy = d sinθ
dy = 8.5 km x sin(30)
dy = 4.25 km
When East and North legs are reversed;dx = 4.25 km
dy = 7.36 km
Resultant displacement;
R = √(dx² + dy²)
R = √(4.25² + 7.36²)
R = 8.5 km
Thus, when the East and North legs are reversed in order, you will still travel 8.5 km.
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suppose the velocity of a particle is given by the equation, V=m+nt2, where m=10cms and n=2cm/s3.
A, find the change in velocity of the particle in the initial time interval b/n t₁=2sec and t₂=5sec
B, find the average acceleration in this time interval
C, find the instantaneous acceleration at time t₁=2sec
Hello!
Begin by plugging in the values for m and n. We get the equation for the velocity of the particle to be:
[tex]v(t) = 10 + 2t^2[/tex]
A.
To find the change in velocity over the interval (2s ≤ t ≤ 5s), we can simply find the difference in the velocities at these times.
[tex]\Delta v= v_f - v_i[/tex]
For this situation:
[tex]\Delta v = v(5) - v(2)[/tex]
Substitute these times for 't' into the equation and solve.
[tex]v(5) = 10 + 2(5^2) = 60 \frac{cm}{s}\\\\v(2) = 10 + 2(2^2) = 18 \frac{cm}{s}\\\\\Delta v = 60 - 18 = \boxed{42 \frac{cm}{s}}[/tex]
B.
To find the average acceleration, we must take the SLOPE of the velocity function over this interval using the slope formula:
[tex]a_{avg} = \frac{v_f - v_i}{\Delta t}[/tex]
Plug in the values for the particle's velocity at t = 2 s and 5 s that we solved for above.
[tex]a_{avg} = \frac{60- 18}{5 - 2}\\\\a_{avg} = \frac{42}{3} = \boxed{ 14 \frac{cm}{s^2}}[/tex]
C.
The instantaneous acceleration can be found by taking the derivative of the v(t) function using the power rule. Recall:
[tex]\frac{dy}{dx} x^n = nx^{n-1}[/tex]
Using this rule:
[tex]a(t) = v'(t) = 2(2t) = 4t[/tex]
Substituting in t = 2 s:
[tex]a(2) = 4(2) = \boxed{8 \frac{cm}{s^2}}[/tex]
Any Acellus folk figure out the answer to this one??? I CANNOT for the life of me.
What is the angular velocity of the minute hand of a clock?
[?] rad/s
Hint: It travels clockwise and takes 60 minutes to make one revolution.
(not 1.25, 1.7, 1800, 30, 60, 0.1047)
Answer:
-0.10472[tex]\frac{rad}{s}[/tex]
Explanation:
Remember that any clockwise rotation is considered negative, and any counterclockwise rotation is considered positive.
The formula for period is:
[tex]T = \dfrac{2\pi}{\omega}[/tex]
Solve for [tex]\omega\\[/tex] and you get:
[tex]\omega=\dfrac{2\pi}{T}[/tex]
The period is 60 seconds so make that division and you're left with -0.10472[tex]\frac{rad}{s}[/tex] or -0.10[tex]\frac{rad}{s}[/tex] if significant figures apply (negative because the rotation is clockwise).
Question 2 of 25
What is the term for a pair of forces described by Newton's third law?
A.
Normal-friction pair
B.
Action-reaction pair
C.
Kinetic-static pair
D.
Inertia-movement pair
The term for a pair of forces described by Newton's third law is action-reaction pair. Details about Newton's law can be found below.
What does Newton's third law state?Newton's third law of motion states that for every action, there is an equal and opposite reaction.
This law proposed that when an object 1 acts on object 2 with a force of a particular magnitude, object 2 also acts on object 1 with an opposite force of same magnitude.
Therefore, it can be said that the term for a pair of forces described by Newton's third law is action-reaction pair.
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Which sentence describes an object that has kinetic energy?
Answer:
Kinetic energy is referred to the energy associated with movement and or motion, thus the only object that is moving would be A boat sailing across the ocean.
Explanation:
On Earth, a scale shows that you weigh 490 N. What is your mass?
A 50 kg
B 100 kg
c) 75 kg
D 22 kg
Answer:
a
Explanation:
Your heart pumps about 70 mL of blood per beat and beats an average of 80 Times per minute how many gallons of blood does your heart pump every hour
Answer:
88.761
Explanation:
