Answer:
M.E = mgh
given, m=400g = 400/1000=0.4kg
h=1.75m
g=9.81m/s²
M.E =0.4×9.81×1.75
=6.867J
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
what is the stress in a steel wire that is 5m long and 0.04cm squared in cross section If the wire bears a load of 20kg?
Answer:
Explanation:
stress = ?
length =5 m
area of cross section = 0.04 cm or 0.0004m
force = 20 × 10 = 200 N ( w = mg) g = 10
formula : stress = force / cross-sectional area
stress = 200 / 0.0004
stress = 500,000 Nm^-2
How does the force of gravity and the force of earth contribute to africa's poverty?
Answer:
The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air
Explanation:
No explanation
I need help with this
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]
On substituting the given values, we get
⇒ [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]
⇒ [tex]=6666.67 \ N[/tex]
6) Which of the following describes a good team member?
A) She is willing to compromise.
B) He is aggressive.
C) She is stubborn.
D) He is conceited.
Answer: A
Explanation:
Because someone who is aggressive, stubborn, or proud of theirselves are more likely to think they're above everyone else and be a bully. However someone who is willing to compromise is better since you can generally make everyone happy that way
HOPE THIS HELPS ^^
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
why is potassium and sodium considered as reactive metals?
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
CiCi is hiking in the woods after a rainstorm when she sees a single large mass of rock and soil moving quickly downhill.
Which type of mass movement is this?
A. landslide
B. slump
C. creep
D. mudflow
Fairly easy question I’ll give extra points help.
1. third law
2. first law
3. third law
4. second law
Pls help me mark Brainliest here the answer choices
4.0N
8.0N
12.0N
16.0N
20.0N
Answer:
20.0N
Becuase It's the largest
Answer:
20.0
Explanation:
It's the biggest number
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
answer asap!!! i suck at acceleration
Answer: 2.67
Explanation: it said he went from 0 to 8 in 3 seconds so if we divide eight By three we get 2.67 rounded to the nearest hundredth so you accelerated that 2.67 m/s
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?
Answer:
The second bulb will have thicker filament
Explanation:
Given;
First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V
Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V
Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m
Resistance of the first bulb:
[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]
Resistance of the second bulb:
[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]
Resistivity of the tungsten filament is given by the following equation;
[tex]\rho = \frac{RA}{L}[/tex]
where;
L is the length of the filament
R is resistance of each filament
A is area of each filament
[tex]A = \pi r^2[/tex]
where;
r is the thickness of each filament
[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]
Therefore, the second bulb will have thicker filament
In which type of circuit does charge move in only one direction?
A. A D.C CIRCUIT
B. AN A.C CIRCUIT
C. A COMBINED CIRCUIT
D. A PARALLEL CIRCUIT
A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Answer:
The speed of the bird is 1.00% of the speed of sound.
Explanation:
The speed of the bird can be found by using the Doppler equation:
[tex] f = f_{0}(\frac{v - v_{r}}{v - v_{s}}) [/tex]
Where:
v: is the speed of sound = 343 m/s
f₀: is the frequency emitted = 1490 Hz
f: is the frequency observed = 1505 Hz
[tex]v_{r}[/tex]: is the speed of the receiver = 0 (it is stationary)
[tex]v_{s}[/tex]: is the speed of the source =?
The minus sign of [tex]v_{s}[/tex] is because the source is moving towards the receiver.
By solving the above equation for [tex]v_{s}[/tex] we have:
[tex] v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s [/tex]
The above speed in terms of the speed of sound is:
[tex]\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%[/tex]
Therefore, the speed of the bird is 1.00% of the speed of sound.
I hope it helps you!
If our atmosphere had a uniform density of 1.25 kg/m3 all the way up to a border with empty space above, that border would be Answer km above sea level. The pressure at sea level is 1 atm = 105 N/m2 and g = 10 m/s2. Enter your answer as an integer.
Answer:
The border is 8km above sea level.
Explanation:
We know that:
Density = 1.25 kg/m^3
Pressure = 10^5 N/m^2
g = 10m/s^2
Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.
This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)
Then the total mass inside that volume is:
M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)
The weight of this mass is:
W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to 10^5 N/m^2
then:
P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
Whit this equation we can find the value of H.
10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
10^5 N = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H
8,000 m^3 = 1m^2*H
(8,000 m^3)/(1m^2) =H
8,000 m = H
And we want this answer in km, knowing that 1,000m = 1km
8,000m = 8km = H
The border is 8km above sea level.
Height of boundaries is 8.2 km
Given that:Normal density = 1.25 kg/m³
1 atm = 101325 N/m²
Find:Height of boundaries
Computation:Pressure = Height × Density × Gravitational acceleration
101325 = Height × 1.25 × 9.8
101325 = Height × 12.25
Height of boundaries = 101325 / 12.25
Height of boundaries = 8271.42 m
Height of boundaries = 8.2 km
Learn more:https://brainly.com/question/23358029
PLEASE ANSWER WITH ACTUAL ANSWER AND I WILL MARK BRAINLIEST (IF YOU GIVE ME A SCAMMY ANSWER I WILL REPORT YOU!!!)
A student wants to determine the local value of the gravitational field strength, g , in their classroom. Which of the following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured?
Select TWO answers.
A: Run a lab cart down an inclined plane; measure the length of the ramp and the time it takes the cart to reach the bottom.
B: Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.
C: Accelerate a lab cart horizontally; measure the mass of the cart and its acceleration.
D: Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.
Answer:
Most likely (B)
Explanation:
B in the passage is the most representative out of all your choices and it has evidence from the passage
Hope dis helps Jit!
Sorry i forgot to type C
B and C both measure mass while the others are calculations and are bias
The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:
Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.What is gravitational field?A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.
When a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.
When a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.
Learn more about gravitational field here:
https://brainly.com/question/26690770
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If a current of 1.10 A flows through a 7.00 Ω resistor of length 3.00 m, what is the electric field strength inside the resistor?
Answer:
the electric field strength inside the resistor is 2.57 V/m
Explanation:
Given;
current flowing through the wire, I = 1.10 A
resistance of the wire, R = 7.00 Ω
length of the wire, L = 3.00 m
The emf created inside the resistor is calculated as;
V = IR
V = 1.10 x 7
V = 7.7 V
The electric field strength inside the resistor is calculated as;
E = V/L
E = 7.7 / 3
E = 2.57 V/m
Therefore, the electric field strength inside the resistor is 2.57 V/m
A child of mass 51.9 kg sits on the edge of a merry-go-round with radius 2.4 m and moment of inertia 215.24 kg m2 . The merry go-round rotates with an angular velocity of 2.1 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.864 m from the center. Now what is the angular velocity of the merry-go-round
Answer:
4.25 rad/s
Explanation:
Given that.
Mass, m = 51.9 kg
Radius, r1 = 2.4 m
Moment of inertia, I = 215.24 kgm^2
Angular velocity, ω = 2.1 rad/s
Radius, r2 = 0.864 m
To start with, we are going to use the Conservation of angular momentum to solve the question, which is
l(initial) = l(final)
[I₁ + I₂](initial)*ω(initial) = [I₁ + I₂](final)*ω(final)
Making ω(final) the subject of formula, we have
ω(final) = [I₁ + I₂](initial)*ω(initial) / [I₁ + I₂](final)
ω(final) = [215.24 + (51.9)(2.4)²](2.1) / [215.24 + (51.9)(0.864)²]
ω(final) = [215.24 + 298.944]2.1 / [215.24 + 38.74]
ω(final) = 514.184 * 2.1 / 253.98
ω(final) = 1079.786 / 253.98
ω(final) = 4.25 rad/s
= 5.273 rad/s
Help please. Question about a potential energy.
Temperature is the measure of the average kinetic energy of the particles in an object. True False
Answer:
true
Explanation:
with increased temperature particles move faster as they gain kinetic energy
Which of the following would NOT be
considered a pollutant?
A. carbon monoxide
B. sulfur dioxide
C. oxygen
D. smoke
Answer:
Answer: Oxygen
Explanation:
Oxygen would not be considered as a pollutant
Answer:
Hey there the answer is C. Oxygen
Explanation:
Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds. After hydrogen and helium, oxygen is the third-most abundant element in the universe by mass. At standard temperature and pressure, two atoms of the element bind to form dioxygen, a colorless and odorless diatomic gas with the formula O 2. Diatomic oxygen gas constitutes 20.95% of the Earth's atmosphere. Oxygen makes up almost half of the Earth's crust in the form of oxides. Hope this helps! Have a great day!
Mass is the amount of matter in an
object. Which statement is true?
A. A person's mass on the moon is less than it is on
Earth.
B. Whether on the Earth or the moon, a person's mass
is the same.
C. Gravity changes the amount of matter there is in an
object.
D. Mass is the same thing as weight.
Each of the two grinding wheels has a diameter of 6 in., a thickness of 3/4 in., and a specific weight of 425 lb/ft3. When switched on, the machine accelerates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec. Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.
Answer:
[tex]0.842\ \text{lb ft}[/tex]
[tex]0.1052\ \text{lb ft}[/tex]
Explanation:
d = Diameter of wheel = 6 in
r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]
t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]
w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]
[tex]t_2[/tex] = Time taken to slow down = 35 s
[tex]t_1[/tex] = Time taken to reach operating speed = 5 s
[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
Weight is given by
[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]
Moment of inertia is given by
[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]
Angular acceleration while slowing down is given by
[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]
Frictional moment is given
[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]
Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]
Angular acceleration while speeding up is given by
[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]
Motor torque is given by
[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]
Motor torque is [tex]0.842\ \text{lb ft}[/tex].
As altitude decreases, what happens to
air pressure?
A. increases
B. decreases
C. stays the same
D. not enough information to tell
Answer:
A. Increases
Explanation:
As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.
Answer:
It decreases so it is B
Explanation:
As altitude rises, air pressure drops.
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.