Answer:
u = 14 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. On the floor just when u speeding out
Em₀ = K = ½ m v²
final point. When on top of the building, no speed
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ m v² = m g h
v = u
u = [tex]\sqrt{2 g h}[/tex]
u= [tex]\sqrt{2 \ 9.8 \ 10}[/tex]
u = 14 m / s
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
Guys please help. I need this question
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
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.
types of wave interactions include
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 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
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
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
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:
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Guys can you please help me real quick with this
Answer:
1. Wavelength = 3.2 m
2. Amplitude = 0.6 m
Explanation:
1. Determination of the wavelength.
The wavelength of a wave is defined as the distance between two successive crest. This implies that for every complete vibration, there is one wavelength.
From the diagram given above, we can see that the wave makes 2½ vibrations.
This means that there are 2½ equal wavelength of the wave. Therefore, the wavelength can be obtained as follow:
Length (L) = 8 m
Wavelength (λ) =?
2½ λ = L
5/2 λ = 8
5λ / 2 = 8
Cross multiply
5λ = 2 × 8
5λ = 16
Divide both side by 5
λ = 16 / 5
λ = 3.2 m
Therefore, wavelength of the wave is 3.2 m.
2. Determination of the amplitude.
The amplitude of a wave is defined as the maximum displacement of the wave from the origin.
From the diagram given above, the distance between the maximum and minimum displacement is given as 1.2 m. Thus, we can obtain the amplitude of wave as follow:
Distance between the maximum and minimum displacement (D) = 1.2
Amplitude (A) =?
A = ½D
A = ½ × 1.2
A = 0.6 m
Thus, the amplitude of the wave is 0.6 m
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)
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
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.
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
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].
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
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:
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?
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].
Are all harmful effects of smoking reversible? Explain your answer.
I need help with this
A bird travels at a speed of 14.2 m/s for 514 meters. How many seconds did it
fly?
Answer:
0.54 sec
Explanation:
Answer:
Time = 36.19 secondsExplanation:
Speed = 14.2 m/s
Distance = 514 m
Time = Distance / Speed
Time = 514 / 14.2
Time = 36.19 seconds
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
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
Night terrors and nightmares are
really the same event.
True
False
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.
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
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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