Answer:
wind orosion is the correct answer dkr this
Which of the following phrases describes a motor?
A. Changes electrical energy to mechanical energy
B. Never used in appliances with moving parts
C. Can be powered by a hand crank
D. Does not use the interaction between a spinning coil of wire and a
magnet
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
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
Simple physics question, check the document. Should take about 3-5 minutes.
Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
What voltage is required to move 6A through 5Ω?
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:
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.
What is gravitonal force
Answer:
its something that hold the air for forceing liy by the exgen
Explanation:
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.
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
A stereo speaker is placed between two observers who are 35 m apart, along the line connecting them. If one observer records an intensity level of 64 dB, and the other records an intensity level of 85 dB, how far is the speaker from each observer
Answer:
x = 2,864 m , Ra = 32.1 m
Explanation:
Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer
observer A β = 64 db
β = 10 log Iₐ / I₀
where I₀ = 1 10⁻¹² W / m²
Iₐ = I₀ 10 (β/ 10)
let's calculate
Iₐ = 1 10⁻¹² (64/10)
Iₐ = 2.51 10⁻⁶ W / m²
Observer B β = 85 db
I_b = 1 10-12 10 (85/10)
I_b = 3.16 10⁻⁴ W / m²
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
P = Ia Aa = Ib Ab
the area of a sphere is
A = 4π R²
we substitute
Ia 4pi Ra2 = Ib 4pi Rb2
Ia Ra2 = Ib Rb2
Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute
Ia (35 -x) 2 = Ib x2
we develop and solve
35-x = Ra (Ib / Ia) x
35 = [Ra (Ib / Ia) +1] x
x (11.22 +1) = 35
x = 35 / 12.22
x = 2,864 m
This is the distance of observer B
The distance from observer A
Ra = 35 - x
Ra = 35 - 2,864
Ra = 32.1 m
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
when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why
Answer:
hsvshxansjusjsnwjwisks
Explanation:
When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..
If a penny is dropped from rest from a tower takes 2 seconds to hit the ground, how far did it travel?
a 29.4 m
b 19.6 m
с 6.8 m
d 9.8 m
Answer:
B
Explanation:
t = 2s
u = 0m/s (released from rest)
a = +g = 9.8m/s²
s = H = ?
using,
s = ut + 1/2at²
H = 0(2) + 1/2(9.8)(2²)
H = 0 + 9.8(2)
s = H = 19.6m
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
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.
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
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]
What is the result of increasing the speed at which a magnet moves in and
out of a wire coil?
A. The current in the wire increases.
B. The magnetic field around the magnet decreases.
C. The current in the wire decreases.
D. The magnetic field around the magnet increases.
Answer:
A. The current in the wire increases.
Explanation:
Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.
This phenomenon shows the inter-relationship between electricity and magnetic fields.
Magnetic fields are induced by passage of electric current. Also, electric current can be produce by magnetic fields. When the speed at which a magnet moves in and out of a wire coil increases, the current also increases.A 0.22 caliber handgun fires a 1.9g bullet at a velocity of 765m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for the bullets?
Answer:
de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ mThe value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.Explanation:
Given;
mass of the bullet, m = 1.9 g = 0.0019 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
[tex]\lambda = \frac{h}{mv}[/tex]
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ is de Broglie wavelength of the bullet
[tex]\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\ \lambda =4.56 *10^{-34} \ m[/tex]
Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Given:
Mass, m = 1.9 g or, 0.0019 kgVelocity, v = 765 mPlank's constant, h = 6.626 × 10⁻³⁴ J/sThe De-Broglie wavelength,
→ [tex]\lambda = \frac{h}{mv}[/tex]
By putting the values,
[tex]= \frac{6.626\times 10^{-34}}{0.0019\times 765}[/tex]
[tex]= 4.56\times 10^{-34} \ m[/tex]
Thus the response above is right.
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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
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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.
This is a short question can anyone help me please
Thank you
Picture Above
Answer:
I thinks it's
deficit spending
Explanation:
cause When a government spends more than it collects in taxes, it is said to have a budget deficit.
F = 5 Newtons
W = 75 Joules
d = ?
ANSWER
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
Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how far south does the arrow travel. (Step 2, you need the previous questions answer to answer correctly
a. 118.4 m south
b 1936 m south
C 2 m south
d 640 m south
Answer:c
Explanation:its the answer because its the answer
An experiment consists of throwing balls straight up with varying initial velocities. Which quantity will have the same value in all trials?a) initial momentum.
b) maximum height.
c) time of travel.
d) acceleration.
Answer:
the correct answer is D, acceleration of gravity
Explanation:
In a projectile launch problem it is described by the expressions
v = v₀ - g t
v² = v₀² - 2 g y
y = v₀ t - ½ g t²
By examining these equations we can see that acceleration is the magnitude that appears constant in all expressions.
This acceleration is the acceleration of gravity with a value of g = 9.8 m/s² and directed towards the center of the Earth
therefore the correct answer is D
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.
A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.05 s?
Answer:
The angular speed is 0.13 rev/s
Explanation:
From the formula
[tex]\tau = I\alpha[/tex]
Where [tex]\tau[/tex] is the torque
[tex]I[/tex] is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
But, the angular acceleration is given by
[tex]\alpha = \frac{\omega}{t}[/tex]
Where [tex]\omega[/tex] is the angular speed
and [tex]t[/tex] is time
Then, we can write that
[tex]\tau = \frac{I\omega}{t}[/tex]
Hence,
[tex]\omega = \frac{\tau t}{I}[/tex]
Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].
Here, The torque is given by,
[tex]\tau = rF[/tex]
Where r is the radius
and F is the force
From the question
r = 3.00 m
F = 195 N
∴ [tex]\tau = 3.00 \times 195[/tex]
[tex]\tau = 585[/tex] Nm
For the moment of inertia,
The moment of inertia of the solid disk is given by
[tex]I = \frac{1}{2}MR^{2}[/tex]
Where M is the mass and
R is the radius
∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]
[tex]I = 1462.5[/tex] kgm²
From the question, time t = 2.05 s.
Putting the values into the equation,
[tex]\omega = \frac{\tau t}{I}[/tex]
[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]
[tex]\omega = 0.82[/tex] rad/s
Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π
0.82 rad/s = 0.82/2π rev/s
= 0.13 rev/s
Hence, the angular speed is 0.13 rev/s,
A 5kg rock is lifted 2m. Find the amount of work done.
Answer:
98J
Explanation:
Given parameters:
Mass of rock = 5kg
Height = 2m
Unknown:
Work done = ?
Solution:
The amount of work done is given as:
Work done = Force x distance
Work done = Weight x height
Work done = mgH
Now insert the parameters and solve;
Work done = 5 x 9.8 x 2 = 98J
The amount of work done on the rock is equal to 98 Nm.
Given the following data:
Mass of rock = 5 kgDistance = 2 metersTo determine the amount of work done:
First of all, we would calculate the force acting on the rock:
[tex]Force = mg\\\\Force = 5 \times 9.8[/tex]
Force = 49 Newton
Now, we can determine the amount of work done:
[tex]Work \;done = force \times distance\\\\Work \;done = 49 \times 2[/tex]
Work done = 98 Nm
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