Answer:
Initial velocity u = 60 ft/s
Explanation:
Given:
Deceleration a = -36 ft/s²
Distance covered s =50ft
Final velocity v = 0 ft/s
Find:
Initial velocity u
Computation:
Using third equation of motion;
v² = u² + 2as
0² = u² + 2(-36)(50)
0 = u² - 3600
u² = 3600
u = 60 ft/s
Initial velocity u = 60 ft/s
If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
Are all harmful effects of smoking reversible? Explain your answer.
A +0.0129 C charge feels a 4110 N
force from a -0.00707 C charge. How
far apart are they?
[?] m
Answer:
r = 14.13 m
Explanation:
Given that,
Charge 1, q₁ = +0.0129 C
Charge 2, q₂ = -0.00707 C
The force between charges, F = 4110 N
We need to find the distance between charges. The formula for the force between charges is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Where
r is the distance between charges
So,
[tex]r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129 \times 0.00707 }{4110 }} \\\\r=14.13\ m[/tex]
So, the distance between charges is equal to 14.13 m.
Answer:
14.13 m
Explanation:
acellus
A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
seconds, what was his acceleration?
Answer:
a = 52s²
Explanation:
How to find acceleration
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Solve
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
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?
Answer:
239 rpm
Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:
100πcm/rev
75,000cm/min
750 min rev≈238.7324146RPM
a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?
Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
if 400g is 1kg find the ratio in the simplest form
2:5
Explanation:
400g : 1kg
400g: 1000g
4 : 10
2 : 5
Balance the equation by choosing the correct coefficient numbers in the drop down menus.
[Select]
SO2 +
[Select]
VH₂ →
[Select]
S +
[ Select]
H20
It is suggested you write this on scratch paper and balance it before choosing your answers :)
Answer:
SO₂ + 2H₂ —> S + 2H₂O
The coefficients are: 1, 2, 1, 2
Explanation:
SO₂ + H₂ —> S + H₂O
The above equation can be balance as follow:
SO₂ + H₂ —> S + H₂O
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:
SO₂ + H₂ —> S + 2H₂O
There are 2 atoms of H on the left side and 4 atoms the right side. It can be balance by writing 2 before H₂ as shown below:
SO₂ + 2H₂ —> S + 2H₂O
Now, the equation is balanced.
The coefficients are: 1, 2, 1, 2
The low pressure area near Earth's equator is filled by cool air moving in from
А
Europe and South America
B
the North and South Pole
с
the Prime Meridian
D
the Atlantic and Pacific Ocean
The volume of a gas decreases from 15.7 mºto 11.2 m3 while the pressure changes from 1.12 atm to 1.67 atm. If the
initial temperature is 245 K, what is the final temperature of the gas?
O 117 K
230 K
261K
.
O 512K
Answer:
Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.
Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.
Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].
In other words:
[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].
[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].
Combine the two equations (equate the right-hand side) to obtain:
[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].
Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:
[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].
[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].
The graph shows the heating curve of water the X axis shows heat added overtime and Y axis shows the temperature identify the regions were liquid water is present
Answer:
liquid, solid, and gas. A heating curve shows how the temperature changes as a substance is heated up at a constant rate.
Explanation:
liquid is often the bridge between solid and gas
not always, but most times.
For water, liquid water would probably be at temperature Y= 32- 212 degrees F, or Y= 0-100 degrees C.
Apologies, I hope this helps.
On the map, which major plate is flanked by the red sea rift and the Minor Arabian Plate?
A:#1 North American Plate
B:#3 South American Plate
C:#5 Eurasian Plate
D:#2 African Plate
Answer:
D:#2 African Plate
Explanation:
The African Plate is flanked by the Red sea rift and the minor African plate.
The Red sea rift is a small part of a greater line of rifts known as the Great African Rift Valley. The rift valley is making several small lakes all through Africa and it will eventually split up the African continent.
The Red sea lift is the divergent boundary between the African plate and the Arabian plate. It means that the two plates are moving apart or spreading apart.
How would increasing the pressure of this reaction affect the equilibrium
Explanation:
c because there is element
Answer:
C. H2 and N2 would react to produce more NH3
Explanation:
A.P.E.X
What is the speed of a ball that is attached to a string and swings in a horizontal circle of radius 2.0 m with the central acceleration of 15 m/s^2?
Answer:
5.48 m/s.
Explanation:
Use the formula a=v^2/r.
An object was thrown from rest upward with an initial velocity of 10m/s with time frame of 6s find the distance of the object from it's resting point
Answer:
60
Explanation:
Work Done= Force×Displacement in the direction of the force
W.D. = 10×6
W.D. = 10×0.6
W.D. = 6m
Which well will give the most water.
YOU WILL GET 50 POINTS
The well that will have most of the water will be well A.
What is an underground water supply?The underground water supply is defined as a type of water that exists underground in saturated zones beneath the land surface.
From the two wells represented in the diagrams above, Well A has water supply from underground which is lacking in well B.
Therefore, well A will have most of the water more than B.
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The photosphere refers to the Sun's:
core
atmosphere
surface
magnetic field
Answer:
The photosphere is the visible "surface" of the sun. So your answer would be C.
Explanation: its right
A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
Answer:
W = 1/2 K x^2
x^2 = 2 * W / K = 2 * 49 J / (N/m) = .218 / m^2
x = .467 m
Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 6.50 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. Part A If the squid has 1.55 kg of water in its cavity, at what speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator
Answer:
10.1 m/s
Explanation:
By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.
And by the law of conservation of momentum,
initial momentum of squid + expelled water = final momentum of squid + expelled water.
Now, the initial momentum of the system is zero.
So, 0 = final momentum of squid + expelled water
0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled
So, MV + mv = 0
MV = -mv
v = -MV/m
= -6.50 kg × 2.40 m/s ÷ 1.55 kg
= -15.6 kgm/s ÷ 1.55 kg
= -10.1 m/s
So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s
PLEASE HELP!! :)
Which of the following options would increase the electric force the most?
a. tripling the charge on one particle
b. changing the sign of one of the particles.
c. doubling the charge on one particle
d. doubling the charge on both particles
Guys please help. I need this question
An object, with mass 64 kg and speed 14 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K_f = 1881.6 J
Explanation:
To solve this exercise, let's start by finding the velocities of the bodies.
We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved
initial instant. Before the explosion
p₀ = M v₀
final instant. After the explosion
p_f = m₁ v + m₂ 0
the moeoto is preserved
p₀ = p_f
M v₀ = m₁ v
v = [tex]\frac{m_1}{M}[/tex] v₀
in the exercise they indicate that the most massive part has twice the other part
M = m₁ + m₂
M = 2m₂ + m₂ = 3 m₂
m₂ = M / 3
so the most massive part is worth
m₁ = 2 M / 3
we substitute
v = ⅔ v₀
with the speed of each element we can look for the kinetic energy
initial
K₀ = ½ M v₀²
Final
K_f = ½ m₁ v² + 0
K_f = ½ (⅔ M) (⅔ v₀)²
K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)
K_f = [tex]\frac{8}{27}[/tex] K₀
the energy added to the system is
ΔK = Kf -K₀
ΔK = (8/27 - 1) K₀
ΔK = -0.7 K₀
K_f = K₀ + ΔK
K_f = K₀ (1 -0.7)
K_f = 0.3 K₀
let's calculate
K_f = 0.3 (½ 64 14²)
K_f = 1881.6 J
a sprinter accelerates from rest to 14m/s in 1.38 s. what is her acceleration in km/h^2
The acceleration of the sprinter is approximately 131,426 km/h^2.
To find the acceleration in km/h^2, we need to convert the units from meters per second (m/s) to kilometers per hour (km/h) and adjust the time units accordingly. Here's the step-by-step calculation:
1. Convert the final velocity from m/s to km/h:
14 m/s * (3.6 km/h) / (1 m/s) = 50.4 km/h
2. Convert the time from seconds (s) to hours (h):
1.38 s * (1 h) / (3600 s) = 0.0003833 h
3. Calculate the acceleration using the formula:
Acceleration = (Final velocity - Initial velocity) / Time
Since the initial velocity is 0 m/s (rest), we have:
Acceleration = (50.4 km/h - 0 km/h) / 0.0003833 h
Acceleration = 131425.955 km/h^2
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What kind of energy is in a moving skateboard
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
What is the force between two 1.0 X 10^-5 C charges separated by 2.0 m?
According to Coulomb's law, the force between the given charges is 0.225N which is explained below.
Coulomb's Law:Force on two identical charges q separate by a distance of r is given by:
F = kq²/r²
where k is Coulomb's constant
q is the charge
r is the separation between the charges
Given that q = 1×10⁻⁵C,
and r = 2m
So, the force between the given charges will be:
F = (9×10⁹)(1×10⁻⁵)²/2²
F = 0.225N is the required force.
Learn more about Coulomb's law:
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Please solve this and please don’t put in a link.
Answer:
synthesis
Explanation:
It is a combination (synthesis) reaction.
A+B→AB
Answer:
i think it is a synthesis
Explanation:
synthesis reaction occurs when two substances combine and form a compound...
hope this helps!!
Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1000 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 60 volts
Answer:
[tex]1.62\times 10^{-8}\ \text{s}[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Vacuum permittivity = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]
[tex]A[/tex] = Area = [tex]10\times 2\times 10^{-4}\ \text{m}^2[/tex]
[tex]d[/tex] = Distance between plates = 1 mm
[tex]V_c[/tex] = Changed voltage = 60 V
[tex]V[/tex] = Initial voltage = 100 V
[tex]R[/tex] = Resistance = [tex]1000\ \Omega[/tex]
Capacitance is given by
[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}[/tex]
We have the relation
[tex]V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}[/tex]
The time taken for the potential difference to reach the required level is [tex]1.62\times 10^{-8}\ \text{s}[/tex].
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
Answer:
a)
the quarterback will be moving back at speed of 0.080625 m/s
b)
the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
Explanation:
Given the data in the question;
a)
How fast will he be moving backward just after releasing the ball?
using conservation of momentum;
m₁v₁ = m₂v₂
v₂ = m₁v₁ / m₂
where m₁ is initial mass ( 0.43 kg )
m₂ is the final mass ( 80 kg )
v₁ is the initial velocity ( 15 m/s )
v₂ is the final velocity
so we substitute
v₂ = ( 0.43 × 15 ) / 80
v₂ = 6.45 / 80
v₂ = 0.080625 m/s
Therefore, the quarterback will be moving back at speed of 0.080625 m/s
b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
we make use of the relation between time, distance and speed;
s = d/t
d = st
where s is the speed ( 0.080625 m/s )
t is time ( 0.30 s )
so we substitute
d = 0.080625 × 0.30
d = 0.0241875 m or 2.41875 cm
Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
1- charging by touch occurs when electrons are transmitted by direct contact.
(Right)
(wrong)
2- Electric charges are preserved, they are not created or destroyed.
(Right)
(wrong)
Answer:
#1 - True (touch) charging when electric conductors actually touch one another.
#2. True - electrical charges are conserved (not destroyed)