step by step
formular v^2 = u^2+2as
stop v = 0
0 = 400+2a(20)
-400=40a
a = -10 m/s^2
ans affect acceleration is 10 m/s^2
the specific heat capacity of a substance is 500J/kg/oC. Find the heat required to rise the temperature of 10 quintial of the substance by 3 degree celcius
[tex]\boxed{\sf Q=mc\Delta T}[/tex]
[tex]\\ \sf\longmapsto Q=1000(5000)(3)[/tex]
[tex]\\ \sf\longmapsto Q=15000000J[/tex]
[tex]\\ \sf\longmapsto Q=1.5\times 10^7J[/tex]
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put
Answer:
300J
Explanation:
Work done = Force x the distance travelled in the direction of the force
=300 x 1
=300J
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
Where does a body have more weight the poor at the eqator of the earth.
Answer:
Explanation:
Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.
Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.
As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball
Answer:
t = 0.437 s
Explanation:
Sound is a wave so its speed is constant
v = x / t
t = x / v
indicates that the distance is x = 150 m
t = 150/343
t = 0.437 s
this is the time it takes to hear the hit
To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
Solving for [tex]v_{0y}[/tex],
[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]
[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]
b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as
[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]
or
[tex]v_{0x} = 150\:\text{m/s}[/tex]
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
explain what would happen if the cell was disconnected from the circuit ( please help me)
Most of the time, "cell" means a battery, and if it gets disconnected, all voltage and current in the circuit goes away.
I know this is a lame answer, but it kinda depends on what the cell was doing in the circuit, and what else is in the circuit besides the cell, and you haven't told us anything about these details, so that's really all we can guess.
Although your question is incomplete a general answer within the concept of your question is provided: when the cell is disconnected from the Circuit the flow of voltage across the circuit will be halted.
A cell/battery is often used as a voltage source in electrical circuits, since the cell is a power source, when the cell is connected to a circuit the cell discharges some of its voltage to the components of the circuit ( such as capacitors ) to keep the circuit functional. so when the cell is disconnected from the circuit the flow of voltage is halted
Hence we can conclude that without a voltage source ( cell ) in a circuit there will be no flow of voltage across the components of the circuit.
learn more : https://brainly.com/question/16598952
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
The question is in the photo.
Answer:
heyaa thereeee
see temperature is rising in interval of
0 to 4 minutes
8 to 10 minutes
but 8 to 10 is NOT in options
so answer is option a) 0 to 4 minutes
:))))
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density
Answer:
velocity is the answer of this question.
Answer:
Velocity is the right answer ok
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
A student has a large number of coins of different diameters, all made of the same metal. She wishes to find the density of the metal by a method involving placing the coins in water.
a) State the formula needed to calculate the density.
b) Describe how the measurements of the required quantities are carried out.
Answer:
a)density = mass /volume
b)to find volume put water into a container .measure the level of water , put the coins into the beaker containing water , measure the level of water again, subtract the new volume withe the first one . the result is the volume of coins
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
find the equivalent resistance of this circuit
Answer:
Req = 564 Ω
Explanation:
The equivalent resistance between R1 and R2:
1/R =1/R1 + 1/R2
1/R =1/960 + 1/640
1/R = 1/384
R = 384
Now, the equivalent resistance between R and R3:
Req = 384 + 180
Req = 564 Ω
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?
Answer:
The charges are + 74.3 μC and - 74.3 μC
Explanation:
Let the charges be q and q'.
Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by
F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m
When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.
They now repel each other.
So, the magnitude of the force of repulsion is given by
F' = k[(q + q')/2][(q + q')/2]/r²
F' = k[(q + q')²/4r²
Since the magnitude of the force of attraction and repulsion are the same, we have that
F = F'
kqq'/r² = k[(q + q')²/4r²
qq' = (q + q')²/4
(q + q')² = 4qq'
q² + 2qq' + q'² = 4qq'
q² + 2qq' - 4qq' + q'² = 0
q² - 2qq' + q'² = 0
(q - q')² = 0
q - q' = 0
q = q'
Substituting q = q' into F, we have
F = kqq'/r²
F = kq²/r²
making q subject of the formula, we have
q² = Fr²/k
q = √(Fr²/k)
q = r√(F/k)
Substituting the values of the variables into the equation, we have
q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)
q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)
q = 0.189 m(0.3923 × 10⁻³ C/m)
q = 0.0743 × 10⁻³ C
q = 74.3 × 10⁻³ × 10⁻³ C
q = 74.3 × 10⁻⁶ C
q = 74.3 μC
Since q and q' initially attract, it implies that they initially had opposite charges.
So, q = 74.3 μC and q' = -74.3 μC
So, the charges are + 74.3 μC and - 74.3 μC
Under normal circumstances: _________
a. Fetal Hb binds to oxygen more tightly than Mb binds.
b. Fetal Hb binds oxygen more tightly in the absence of 2,3-BPG.
c. Fetal Hb does not bind to oxygen.
d. Adult Hb has the lowest affinity for oxygen of the 3.
e. More than one of these statements is correct.
Answer:
Fetal Hb binds oxygen more tightly than adult Hb (not option a)
an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?
Answer:
Explanation:
Given force between 2 currents carrying
wires = F₀
Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L
where I₁=Current in wire 1
I₂= Current in wire 2
L= Length of the wire
when one current is doubled and the other is halved
I₁= 2 I₁
I₂= I₂/2
F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L
name a device that converts mechanical energy into electrical energy.
Answer:
Electric generator is the device that converts mechanical energy into electrical energy
