Answer:
a = 7.75 [m/²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}[/tex]
where:
x = final distance = 190 [m]
Xo = initial distance = 0
Vo = initial velocity = 0 (car starts from the rest)
a = acceleration [m/s²]
t = time = 7 [s]
190 = 0 + (0*7) + 0.5*a*(7²)
190 = 0.5*49*a
a = 7.75 [m/²]
In the video your blood is compared to a __________________ that delivers oxygen to your body and picks up CO2 to be released out when you breath.
Answer:
delivery truck
Explanation:
because i got it right
I have a pen
I have a apple
what do I have now?
Answer:
You have an apple pen. :)
Answer:
I have a pen
I have a apple
apple pen
Explanation:
An unbalanced 16.0N force is applied to a2.0kg mass. What is the acceleration of the mass?
Answer:
Yuh
Explanation:
In a sound wave, the wavelength is equivalent to the distance from a region of high pressure to the region of mean
pressure
True
False
What is Newton's universal law of gravitation
Answer:
an object that is in motion wont go out of motion until there is another force pushing on it
which thermometer is used in hot region.why?
Answer:
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.
Explanation:
please mark me brainlist
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.
Four balls have the same temperature. Which ball has the most thermal
energy?
A. Golf ball
B. Bowling ball
C. Tennis ball
D. Basketball
Answer:
a bowling ball because it has the most mass.
Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg
Answer:
m = 4.9 10⁸ kg
Explanation:
The expression for the density is
ρ = m / V
m = ρ V
the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant
V = V_atmosphere - V_planet
V = 4/3 π R_atmosphere³ - 4/3 π R_venus³
V = 4/3 π (R_atmosphere³ - R_venus³
)
the radius of the planet is R_venus = 6.06 10⁶ m.
The radius of the outermost layer of the atmosphere
R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶
R_atmosphere = 6.11 10⁶ m
let's find the volume
V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]
V = 23,265 10⁶ m³
let's calculate the mass
m = 21 23,265 10⁶
m = 4.89 10⁸ kg
with two significant figurars is
m = 4.9 10⁸ kg
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The other end of the spring is fixed to a wall. The spring is compressed by 10.0 cm from its equilibrium position and released from rest.
1) What is the speed of the object when it is 8.00 cm from equilibrium? (Express your answer to three significant figures.)
2) What is the speed when the object is 5.00 cm from equilibrium? (Express your answer to three significant figures.)
3) What is the speed when the object is at the equilibrium position? (Express your answer to three significant figures.)
Answer:
1) v = 0.45 m/s
2) v = 0.65 m/s
3) v = 0.75 m/s
Explanation:
1) We can find the speed of the object by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
Where:
k: is the spring constant = 280 N/m
v: is the speed of the object =?
m: is the mass of the object = 5.00 kg
x: is the displacement of the spring
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]
2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]
3) When the object is at the equilibrium position, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]
I hope it helps you!
(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s
(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s
(3) the speed of the object when the spring is at equilibrium is 0.748 m/s
Compression of spring and conservation of energy:
Given that the mass of the object, m = 5 kg
spring constant, k = 280 N/m
compression of the spring , x = 10 cm = 0.1m
(i) the spring compression is at d = 8cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]
v = 0.449 m/s
(ii) the spring compression is at d = 5cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]
v = 0.648 m/s
(iii) the spring is at equilibrium so compression is at d = 0cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]
v = 0.748 m/s
Learn more about conservation of energy:
https://brainly.com/question/14245799?referrer=searchResults
A crate is pulled due south with a force of 350. N. What other force must be applied if the
net force on the crate is 425 N due north? Enter the magnitude (with units) and direction
(north, south, east, west).
Answer:
775 N due North.
Explanation:
If the crate is pulled South with 350 N force, and the net force on the crate results into 425 N due North, then the other force (F) acting must be larger than the 350 N, and pointing North:
F - 350 N = 425 N
F = 425 N + 350 N = 775 N due North.
What is gravitonal force
Answer:
its something that hold the air for forceing liy by the exgen
Explanation:
The components of vector Upper A Overscript right-arrow EndScripts are Ax and Ay (both positive), and the angle that it makes with respect to the positive xaxis is θ. Find the angle θ if the components of the displacement vector Upper A Overscript right-arrow EndScripts are:
(a) Ax = 12 m and Ay = 12 m,
(b) Ax= 19 m and Ay = 12 m, and
(c) Ax = 12 m and Ay = 19 m.
(a) θ = Number____________ Units____
(b) θ = Number____________ Units____
(c) θ = Number ____________Units____
Answer:
(a) θ = 45° = 0.78 rad
(b) θ = 32.27° = 0.56 rad
(c) θ = 57.27° = 1 rad
Explanation:
When a vector is resolved into its rectangular components, the formula for the direction angle of the vector with positive x-axis is given as:
tan θ = Ay/Ax
θ = tan⁻¹(Ay/Ax)
(a)
Ax = 12 m
Ay = 12 m
θ = tan⁻¹(12 m/ 12 m)
θ = tan⁻¹(1)
θ = 45° = 0.78 rad
(b)
Ax = 19 m
Ay = 12 m
θ = tan⁻¹(12 m/19 m)
θ = tan⁻¹(0.6315)
θ = 32.27° = 0.56 rad
(c)
Ax = 12 m
Ay = 19 m
θ = tan⁻¹(19 m/12 m)
θ = tan⁻¹(1.58333)
θ = 57.27° = 1 rad
An eraser is thrown upward with an initial velocity of 5.0m/s. The eraser’s velocity after 7.0 second is
Answer:
-63.6m/s
Explanation:
Given parameters:
Initial velocity = 5m/s
Time of flight = 7s
Unknown:
Velocity of the eraser after 7s = ?
Solution:
To solve this problem, we have to use the right motion equation which is given below;
v = u - gt
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
t is the time taken;
Now insert the parameters and solve for v;
v = 5 - (9.8 x 7)
v = -63.6m/s
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons is this charge?
Answer:
[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]
Explanation:
Coulomb's Law
The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:
[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]
Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:
[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]
Substituting values:
[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]
[tex]q = 1.05\cdot 10^{-5}~c[/tex]
This charge corresponds to a number of electrons given by the elementary charge of the electron:
[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]
Thus, the charge of any of the spheres is:
[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]
[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]
A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?
A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s
Answer:
B. 1275 kg*m/s
Explanation:
I = F(deltaT) = (deltaP) = mv2- mv1
Therefore,
I = mv2-mv1
m = 850 kg
v2 = 5 m/s
v1 = 3.5 m/s
I = (850)(5)-(850)(3.5)
I = 1275 kg* m/s
A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?
Answer:
x = 1127 [m]
Explanation:
In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.
[tex]v_{f} =v_{o} -a*t[/tex]
where:
Vf = final velocity = 9 [m/s]
Vo = initial velocity = 55 [m/s]
a = acceleration o desacceleration [m/s²]
t = time = 49 [s]
Now replacing:
9 = 55 - a*49
a*49 = 55 + 9
a = 1.306 [m/s²]
Note: The negative sign in the above equation means that the speed decreases.
Now using the second equation.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
(9)² = (55)² - 2*(1.306)*x
2944 = 2.612*x
x = 1127 [m]
The purpose of a motor is to:
A. convert electrical energy to mechanical energy.
B. provide a safe circuit for current flow.
C. convert electrical energy to nuclear energy.
D. convert chemical energy to electrical energy.
Answer:
a motor is used to covert electrical energy to mechanical energy
How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
Explanation:
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
Zinc has a work function of 4.3 eV. a. What is the longest wavelength of light that will release an electron from a zinc surface? b. A 4.7 eV photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron?
Answer:
a
[tex]\lambda_{long} = 288.5 \ nm[/tex]
b
The velocity is [tex]v = 3.7 *0^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The work function of Zinc is [tex]W = 4.3 eV[/tex]
Generally the work function can be mathematically represented as
[tex]E_o = \frac{hc}{\lambda_{long}}[/tex]
=> [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]
Here h is the Planck constant with the value [tex]h = 4.1357 * 10^{-15} eV s[/tex]
and c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]
=> [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]
=> [tex]\lambda_{long} = 288.5 \ nm[/tex]
Generally the kinetic energy of the emitted electron is mathematically represented as
[tex]K = E -E_o[/tex]
Here E is the energy of the photon that strikes the surface
So
[tex]E- E_o = \frac{1}{2} m * v^2[/tex]
Here m is the mass of electron with value [tex]m = 9.11*10^{-31 } \ kg[/tex]
Generally [tex]1 ev = 1.60 *10^{-19} \ J[/tex]
=> [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]
=> [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]
=> [tex]v = 3.7 *0^{5} \ m/s[/tex]
Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat of 1,700 J/(kg°C).
Answer:
Q = 425 kJ
Explanation:
Given that,
Mass, m = 25 kg
The clown doll head that heats up from 25°C to 35°C
The specific heat is 1700 J/kg°C
We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :
[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]
So, 425 kJ of thermal energy is severed.
Describe and give an example of mutualism.
Describe and give an example of commensalism.
Describe and give an example of parasitism.
Describe and give an example of competition.
Describe and give an example of predation.
Answer:
Mutualism, commensalism, parasitism, competition, and predation.
Explanation:
mutualism- relationship between two or more organisms where both are benefited. Example-oxpecker with rhino/zebra. They eat bugs off of them which means that they are getting food, while the rhino/zebra are getting cleaned up with pest control.
commensalism- relationship between two organisms where one benefits and the other isn't benefited or harmed. EX- tree frogs use plants as protectioin.he frog is benefited, and the plant is neither harmed nor benefited. Remora fish have a disk on their heads that they use to attach themselves to larger animals for protection. The animals they attach to are neither harmed nor benefited.
parasitism- in a relationship where an organism benefits at the expense of the other. (one is benefited while the other is harmed) ex- fleas and ticks that live on cats and dogs, tape worms that live in people and animals that eat the food which means that the people aren't getting the food or nutrition that they eat. lice, etc
competition- interaction within organisms/species in which both the organisms/species are harmed and is apart of natural selection. Examples may include two males fighting over a mate, animals competing over food, limited habitats that they are fighting over, territory, etc.
predation- the preying of one animal on another. It's where the predator hunts and eats another organism which is its prey. categorized within-(1) carnivory, (2) herbivory, (3) parasitism, and (4) mutualism. Each type of predation can by categorized based on whether or not it results in the death of the prey.ex- owls hunting mice, wolves hunting rabbits, lion hunting gazelle, etc.
38. You are fishing and catch a fish with a mass of
6kg. If the fishing line can withstand a maximum
tension of 30 N, what is the maximum acceleration
you can give the fish as you reel it in?..*
(10 Points)
Enter your answer
Answer:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Tension of the line = the weight + force from pulling up the fish
30 N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6 a
a = 1.7333333 m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
To learn more about acceleration refer to the link:
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A container is filled to a depth of 21.0 cm with water. On top of the water floats a 35.0-cm-thick layer of oil with specific gravity 0.600. What is the absolute pressure at the bottom of the container
Answer:
P_abs = 105120.2 N/m²
Explanation:
We are given;
Specific gravity of oil; ρ_oil = 0.6 g/cm³ = 600 kg/m³
Depth of water; h_w = 21 cm = 0.21 m
Depth of oil; h_o = 35 cm = 0.35 m
From tables specific gravity of water is; ρ_w = 1000 kg/m³
Thus, to get the absolute pressure at the bottom of the container, we will use the formula;
P_abs = (ρ_w × g × h_w) + (ρ_oil × g × h_oil) + P_a
Where P_a is atmospheric pressure with a standard value of 1.01 × 10^(5) N/m²
g is gravitational acceleration = 9.81 m/s²
Thus;
P_abs = (1000 × 9.81 × 0.21) + (600 × 9.81 × 0.35) + (1.01 × 10^(5))
P_abs = 105120.2 N/m²
F = 5 Newtons
W = 75 Joules
d = ?
ANSWER
Air enters into the hollow propeller tube at A with a mass flow of 4 kg/s and exits at the ends B and C with a velocity of 400 m/s, measured relative to the tube. If the tube rotates at 1500 rev/min, determine the frictional torque M on the tube.
Answer:
643N.m
Explanation:
From this question we have:
Mass flow = 4kg/s
Velocity V = 400m/s
Rotation N = 1500rev/min
We get the relative velocity at exit to be:
V2 = V - r2w
400-0.5x [(2*π*1500)/60]
= 400-78.5
= 321.5m/s
Then we have to calculate the frictional torque My
Mt = Mr2 x V2
= 4x0.5x321.5
= 643Nm
From the calculations above, we get the frictional torque M on the tube to be 643Nm.
What is the average power consumption in watts of an appliance that uses 4.69 kW · h of energy per day?
Answer:
The average power is [tex]P = 195 .42 \ W[/tex]
Explanation:
From the question we are told that
The energy of the appliance is [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]
The time considered is [tex]t = 1 \ day = 24 \ hours[/tex]
Generally the average power consumption is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
=> [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]
=> [tex]P = 195 .42 \ W[/tex]
What horizontal speed must a pumpkin be thrown to hit a car 13.4 meters away from a building which stands 10.4 meters tall?
A) 1.5 m/s
B) 2.1 m/s
C)6.1 m/s
D) 8.9 m/s
Answer:
V₀ₓ = 9.2 m/s
Nearest answer:
D) 8.9 m/s
Explanation:
First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:
h = V₀y t + (1/2)gt²
where,
h = height of building = 10.4 m
V₀y = vertical component of initial speed = 0 m/s
t = time = ?
g = 9.8 m/s²
Therefore,
10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²
t² = (10.4 m)(2)/(9.8 m/s²)
t = √[2.122 s²]
t = 1.45 s
Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,
s = V₀ₓ t
where,
s = horizontal distance between building and car = 13.4 m
V₀ₓ = Horizontal Component of Initial Velocity = ?
Therefore,
13.4 m = V₀ₓ(1.45 s)
V₀ₓ = 13.4 m/1.45 s
V₀ₓ = 9.2 m/s
which is not a type of mechanical wave?
Please help!How is constant or uniform acceleration used to explain free fall?
Use the information below for the next five questions:
An open organ pipe emits B (494 Hz) when the temperature is 14°C. The speed of sound in air is v≈(331 + 0.60T)m/s, where T is the temperature in °C
Determine the length of the pipe.
What is the wavelength of the fundamental standing wave in the pipe?
What is frequency of the fundamental standing wave in the pipe?
What is the frequency in the traveling sound wave produced in the outside air?
What is the wavelength in the traveling sound wave produced in the outside air?
How far from the mouthpiece of the flute should the hole be that must be uncovered to play D above middle C at 294 Hz? The speed of sound in air is 343 m/s.
Answer: Please see answer in explanation column.
Explanation:
Given that
v≈(331 + 0.60T)m/s
where Temperature, T = 14°C
v≈(331 + 0.60 x 14)m/s
v =331+ 8.4 = 339.4m/s
In our solvings, note that
f= frequency
λ=wavelength
L = length
v= speed of sound
a) Length of the pipe is calculated using the fundamental frequency formulae that
f=v/2L
Length = v/ 2f
= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m
b) wavelength of the fundamental standing wave in the pipe
L = nλ/2,
λ = 2L/ n
λ( wavelength )= 2 x 0.3435/ 1
= 0.687m
c) frequency of the fundamental standing wave in the pipe
F = v/ λ
= 339.4m/s/0.687m=
494.03s^-1 = 494 Hz
d) the frequency in the traveling sound wave produced in the outside air.
This is the same as the frequency in the open organ pipe = 494Hz
e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m
f) To play D above middle c . the distance is given by
L =v/ 2 f
= 343/ 2 x 294
=0.583m