Answer:
V₁ = 1.75 m³
Explanation:
Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
where,
P₁ = Initial Pressure of the Gas = 4 KPa
V₁ = Initial Volume of the Gas = ?
P₂ = Final Pressure of the Gas = 2 KPa
V₂ = Final Volume of the Gas = 3.5 m³
Therefore,
[tex](4\ KPa)V_{1} = (2\ KPa)(3.5\ m^{3})\\\\V_{1}=\frac{2\ KPa}{4\ KPa}(3.5\ m^{3})\\\\[/tex]
V₁ = 1.75 m³
____ and____ are 2 major atmospheric gases
Answer:
Nitrogen and Oxygen are the two major atmospheric gases.
The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000-lb car B is traveling at 6 ft/s to the left. If the cars crash head-on, and at a time instant during the crash the impact force on A is 900 lb to the left, what is magnitude and direction of the impact force exerted on B at the instant
Answer:
900 lb to the right
Explanation:
Newton's third law of motion states that for every action there is equal and opposite reaction.
Hence if the two cars (i.e. car A and car B) crash head on and there is an impact on car A of 900 lb to the left, car B would have to generate a force of equal magnitude and in an opposite direction. Since car A had a force of 900 lb to the left, the impact force exerted on car B would be 900 lb to the right.
The answer is 900 lb to the right
Which characterictic of motion could change without changing the velocity of an object
Answer:
The direction could change
Solve ez points. I cant rlly get but i guess most of u do, so its ez
Answer:
A: nucleus
D: neutron
hope this helps :D
Answer:
A: nucleus
D: neutron
Explanation: hope this helps! <3
what is the twisted ladder shape of the DNA called?
Answer:
Double helix
Explanation:
The Double helix is a DNA molecule. The two strands around the Double Helix is called the twisted ladder.
Answer:
double helix
Explanation:
Please help with an explanation for 30 points even if u only do one (but with good explanation)
Answer:
1- 4Ω
Explanation:
if 2Ω is 2 meters, than 4 metres is 4Ω
A boy throws a stone straight upward with an initial
speed of 17.0 m/s.
Part A
What maximum height will the stone reach before falling back down?
Answer:
14.8m
Explanation:
Given parameters:
Initial speed = 17m/s
Unknown:
Maximum height = ?
Solution:
At the maximum height, the final speed will be 0m/s;
We use of the kinematics equation to solve this problem.
V² = U² - 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height
0² = 17² - (2 x 9.8 x h )
0 = 289 - (9.6h)
-289 = -19.6h
h = 14.8m
Which of the following is NOT a type of acceleration?
O A.
Positive acceleration
B.
Constant velocity
O c.
Change in direction
D.
Negative acceleration
The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.4 m/s in the positive x direction and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
33.5J
Explanation:
Given:
Mass of the canister= 3.1 kg
Initial velocity of canister= v(i)= 4.4i m/s
Final velocity of canister= v(f)= 6.4j m/s
Force magnitude( xy plane)= 5 N
The magnitude of vector V'= Vxi + Vyj + Vzk
|V|= √( Vx^2 + Vy^2 + Vz^2
From Kinectic energy and work theorem.
Net work = Kinectic energy of the canister
ΔK= W
(Kf - Ki)= W
Where Kf= final Kinectic energy
= 1/2 mv^2
If we input the given values we have,
= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J
Ki= initial Kinectic energy
= 1/2 mv^2
If we substitute the given values we have
=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J
Work done by canister = (final Kinectic energy - initial energy)
= 63.5- 30
=33.5J
Hence, work done on the canister 33.5J
The velocity of the transverse waves produced by an earthquake is 5.09 km/s, while that of the longitudinal waves is 8.5512 km/s. A seismograph records the arrival of the transverse waves 64.9 s after that of the longitudinal waves. How far away was the earthquake
Answer:
Explanation:
Velocity of transverse wave Vt = 5.09 km/s
Velocity of longitudinal wave Vl = 8.5512 km/s
Let earthquake occurred d km away .
Time taken by transverse wave to travel d distance
= d / 5.09 s
Time taken by longitudinal wave to travel d distance
= d / 8.5512 s
According to question
[tex]\frac{d}{5.09} -\frac{d}{8.5512} = 64.9[/tex]
.19646 d - .11694 d = 64.9
.07952 d = 64.9
d = 816.15 km .
a) Two workers are trying to move a heavy crate. One pushes onthe crate with a force A, which has amagnitude of 445 newtons and is directed due west. The other pusheswith a force B, which has a magnitude of 325newtons and is directed due north. What are the magnitude anddirection of the resultant force A + Bapplied to the crate?
b) Suppose the second worker applies a force of-B instead of B. What then arethe magnitude and direction of the resultant force A -B applied to the crate? In both cases express thedirection relative to due west.
Answer:
Divide then multiply or multiply then divide
Explanation:
to get the answer of a and b
NEED ASAP!! A box of mass 10 kg requires 20 N to slide it across a surface. What is the weight of the box? What is the coefficient of friction between the box and surface ?
Answer: 7
Explanation: 7 is the superior number
can you guys help me to solve first question????
Torque question
A uniform spherical shell of mass M = 11.0 kg and radius R = 0.480 m can rotate about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.160 kg m2 and radius r = 0.110 m, and is attached to a small object of mass m = 1.60 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen a distance 0.700 m after being released from rest? Use energy considerations.
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the charge Q is distributed uniformly, in the shell between r1 and r2, determine the electric field, magnitude and direction, in the following situations:______.
a. From r=0 to r=r1.
b. From r=r1 to r=r2.
c. Outside of r=r2.
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
Suppose you have two metal spheres that are exactly the same size, separated by a very large distance. Sphere A carries an excess of eight negative charges while sphere B carries an excess of two positive charges. Sphere A is momentarily connected to Sphere B using a metal wire and then the wire is removed. Draw charge diagrams of the two spheres for the two cases below before and after the connecting wire is used.
1. Before the spheres are connected using the wire .
2. After the wire has been removed .
Answer:
a) phere A has 8 negative charges and sphere B has 2 positive charges
b) each one has then 3 negative charges
Explanation:
In this case, it is asked to determine the charge of the spheres in two conditions
a) Before connecting the cable, sphere A has 8 negative charges and sphere B has 2 positive charges
b) After connecting the cable, as the spheres are metallic, the load is distributed, we have
q = 8 -2 = 6 negative charges
these charges are distributed between the two spheres, each one has then 3 negative charges
Which property of water causes the cracks in the pavement in cold climates ?
“As temperatures drop, the pavement contracts, building up tensile stresses that lead to cracking,” states MnDOT's Research Services Section. “Fractures occur every 20 to 30 feet across the lane, allowing water to penetrate the structure, which further weakens the pavement layer and the base beneath
Bruce pulls a spring with a spring constant k=100 Nmk=100\, \dfrac{\text N}{\text m}k=100mNk, equals, 100, start fraction, start text, N, end text, divided by, start text, m, end text, end fraction, stretching it from its rest length of 0.20 m0.20\,\text m0.20m0, point, 20, start text, m, end text to 0.40 m0.40\,\text m0.40m0, point, 40, start text, m, end text.What is the elastic potential energy stored in the spring?
Answer:
K_{e} = 2.0 J
Explanation:
In this exercise you are asked to calculate the elastic potential energy of a spring
[tex]K_{e}[/tex] = ½ k x²
where k is the spring constant and x is the displacement from equilibrium position
In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation
Δx = x₁ - x₀
Δx = 0.40 - 0.20
Δx = 0.20 m
let's calculate the elastic potential energy
K_{e} = ½ 100 0.20²
K_{e} = 2.0 J
What is Nature? How does it explain the human condition?
Armand is monitoring a large sealed tank by way of a sensor that records the liquid level over time on a graph. He looks at the graph and claims that the sensor indicates there are waves in the liquid in the tank.
In the space provided, answer each of the following.
Part A: Explain how Armand knows that there is a wave in the tank.
Part B: Find the amplitude, in centimeters (cm), and frequency, in number of waves every second/cycle per second (Hz), of the wave.
Answer:
A.) armand probably looked at the graph that the liquid sensor sends information too
Explanation:
A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0260 kg, and its speed is 42.8 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = [tex]p_{fx} - p_{ox}[/tex]
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m [tex]v_{fx}[/tex] - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
[tex]\Delta p_{y}[/tex] = 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
A 50kg person climbs a 2-meter ladder in 5 second. What is their power output in Watts as they do so?
Answer:
The answer is 196
Explanation:
Because the on the test I got it right
Their power output in watts as they do so will be "200 W". To understand the calculation, check below.
Power and Work done:According to the question,
Person's mass, m = 50 kg
Distance, d = 2 m
Time, t = 5 seconds
As we know,
Force = Mass × Acceleration
= 50 × 10
= 500 N
Now,
Work done = Force × Distance
= 500 × 2
= 1000 J
hence,
The Power will be:
= [tex]\frac{Work \ done }{Time}[/tex]
By substituting the values,
= [tex]\frac{1000}{5}[/tex]
= 200 W
Thus the above answer is correct.
Find out more information about power here:
https://brainly.com/question/1634438
The pendulum on a grandfather clock is 0.993 m long and swings to a maximum 4.57° angle. If the bob of the pendulum has mass = 0.415 kg, how much PE does it have at the top of its swing?
Answer:
0.0128
Explanation:
If the speed of an object is doubled, what happens to its kinetic energy?
Answer:
Kinetic energy is energy of motion. Doubling the speed will quadruple the kinetic energy. The relationship between speed and kinetic energy is: This means that the factor by which kinetic energy increases is the square of the factor by which speed or velocity increases for a given object.
Explanation:
Which is not a product of coal?
A. Monoxide
B. Carbon Dioxide
C. Methane
D. Calcium
Answer:
D
Explanation:
Hope this helps
calculated the pressure exerted when a force of 100N is applied on a area of 25m2
Answer:
P = 4[Pa]
Explanation:
Pressure is defined as the relationship between Force over area. So we can use the following equation.
[tex]P=F/A[/tex]
where:
P = pressure [Pa] (units of Pascals)
F = force = 100 [N]
A = area = 25 [m²]
Now replacing:
[tex]P=100/25\\P=4[Pa][/tex]
[tex]\\ \rm\longmapsto P=\dfrac{F}{A}[/tex]
[tex]\\ \rm\longmapsto P=\dfrac{100}{25}[/tex]
[tex]\\ \rm\longmapsto P=4Pa[/tex]
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m. It climbs a frictional incline of height 2.9 m inclined at an angle 16°, then moves along a second frictional surface of coefficient 0.1 before coming to rest.
The acceleration of gravity is 9.8 m/s^2. If the first frictional surface has a coefficient of 0.21 for a distance 1 m, how far does it slide along the second frictional region before coming to rest?
Answer:
[tex]D=99.4665307m \approx 99.5m[/tex]
Explanation:
From the question we are told that
Mass [tex]m=6kg[/tex]
Velocity of mass [tex]V_m=16[/tex]
Force of Tunnel [tex]F_t=8N[/tex]
Length of Tunnel [tex]L_t=1.6[/tex]
Height of frictional incline [tex]H_i=2.9[/tex]
Angle of inclination [tex]\angle =16 \textdegree[/tex]
Acceleration due to gravity [tex]g=9.8m/s^2[/tex]
First Frictional surface has a coefficient [tex]\alpha_1 =0.21\ for\ d_c=1[/tex]
Second Frictional surface has a coefficient [tex]\alpha _2=0.1[/tex]
Generally the initial Kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}(6)(16)^2[/tex]
[tex]K.E=768[/tex]
Generally the work done by the Tunnel is mathematically given as
[tex]w_t=F_t*d_t[/tex]
[tex]w_t=8*1.6[/tex]
[tex]w_t=12.8J[/tex]
Therefore
[tex]Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t[/tex]
[tex]E_t=K.E+E_t\\E_t=768J+12.8J[/tex]
[tex]E_t=780.8J[/tex]
Generally the energy lost while climbing is mathematically given as
[tex]E_c=mgh[/tex]
[tex]E_c=(6)(9.8)(2.9)[/tex]
[tex]E_c=170.52J[/tex]
Generally the energy lost to friction is mathematically given as
[tex]E_f=\alpha *m*g*cos\textdegree*d_c[/tex]
[tex]E_f=0.21*6*9.8*cos16*1[/tex]
[tex]E_f=11.86965942 \approx 12J[/tex]
Generally the energy left in the form of mass [tex]Em[/tex] is mathematically given as
[tex]E_m=E_t+E_c+E_f[/tex]
[tex]E_m=(768J)-(170.52)-(12)[/tex]
[tex]E_m=585.48J[/tex]
Since
[tex]E_m=\alpha_2*g*m*d[/tex]
Therefore
It slide along the second frictional region
[tex]D=\frac{585.46}{0.1*9.81*6}[/tex]
[tex]D=99.4665307m \approx 99.5m[/tex]
List two things affected by friction PLEASE RESPOND I NEED HELP
metal and hair I think hope this helps
Answer:
cars and trains
Explanation:
i hope that good
Please helpp this is due in 10 minutes?!!!!
Describe the potential energy and kinetic energy at each position of the
roller coaster car. Use these questions to help you.
1.Where is potential energy at its maximum?
2.Where is kinetic energy at its maximum?
3.Where is potential energy increasing and decreasing?
.4Where is kinetic energy increasing and decreasing?
Answer:
1. at the top of the coaster.
2. at the bottom of the coaster.
3. when the car is moving
4. when the car is moving
Explanation:
there is the most amount if potential energy at the top, and the keast at the bottom.
An easy way to draw a circle is to hold a string fixed at one end (for instance, with a pin) and attach a pen to the other end. Holding the string taut and drawing wherever the taut string allows gives a circle. This is the same technique you follow when using a compass. With a compass, the string is simply replaced by a metal or plastic structure, which usually has some markings to let you pick the radius of your circle. The radius of a circle is the only measure that you need to determine any other measure, such as the circumference or area. For instance, the circumference C of a circle (the length measured around the outside of the circle, i.e., the perimeter) is C=2πr. Suppose that you have a piece of string 7 cm long. If one end is held fixed and you draw with a pen at the other end, keeping the string tight, then you will draw a circle.
Required:
a. What is the circumference of this circle?
b. Suppose that you needed to make a pen for some small animals. You have 12 m of fencing. You decide to make a circular pen, because if you wish to enclose an area using a given length of fencing, then a circular fence encloses a larger area than a fence of any other shape. What is the radius r of the pen?
Answer:
The answer is "44 cm and 1.910 m".
Explanation:
In point a:
Formula for Circumference [tex](C)=2 \pi r[/tex]
[tex]= 2 \times 3.14 \times 7 \ (cm)\\\\= 14 \times 3.14 \ (cm)\\\\=43.96 \ cm\\\\=44 \ cm\\[/tex]
In point b:
The formula for the radius of the pen:
[tex]\to r= \frac{C}{2\pi}[/tex] [tex]\therefore C= 12\\[/tex]
[tex]= \frac{12\ m}{2(3.14)}\\\\= \frac{12\ m}{6.28}\\\\= 1.910 \ m \\\\[/tex]
Which is the best tool to use when measuring mass
Beam balance is used to determine the mass of an object.
