A 2.00 kg rock is dropped from the top of a 30.0 m high building. Calculate the ball’s momentum at the time that it strikes the ground.
Explanation:
We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:
Ei = Ef
Ki + Ui = Kf + Uf
Ui = Kf
m g h = 1/2 m v^2
v = sqrt(2gh)
So the momentum will be:
p = mv = m * sqrt(2gh)
Make a generalization about Earth's magnetic field and its properties. In not more than 10 sentences.
Answer:
Earth's magnetic field (and the surface magnetic field) is approximately a magnetic dipole, with the magnetic field S pole near the Earth's geographic north pole (see Magnetic North Pole) and the other magnetic field N pole near the Earth's geographic south pole (see Magnetic South Pole). This makes the compass usable for navigation. The cause of the field can be explained by dynamo theory. A magnetic field extends infinitely, though it weakens with distance from its source. The Earth's magnetic field, also called the geomagnetic field, which effectively extends several tens of thousands of kilometres into space, forms the Earth's magnetosphere. A paleomagnetic study of Australian red dacite and pillow basalt has estimated the magnetic field to be at least 3.5 billion years old
Hey guys can you help me solve this problem "how long will it take a car travelling 30m/s to come to stop ifs its acceleration is -3 m/s2".
Answer:
10 seconds.
Explanation:
We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time t:
[tex]\displaystyle v_f = v_i + at[/tex]
The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².
Substitute and solve for t:
[tex]\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\ t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}[/tex]
Hence, it will take the car 10 seconds to come to a stop.
A wire that is 1.0 m long with a mass of 90 g is under a tension of 710 N. When a transverse wave travels on the wire, its wavelength is 0.10 m. What is the frequency of this wave?
Answer:
890 HzI hope you liked my answer. Thank You!
Use the equation of motion to answer the question. Use the equation of motion to answer the question.
x=x0+v0t+12at2
An object has a starting position of x = 2 m, a starting velocity of 4.5 m/s, and no acceleration. Which option shows the final position of the object after 2 s?
The final position of the object after 2 s is 11 m.
Motion: This can be defined as the change in position of a body.
⇒ Formula:
x = x₀+v₀t+1/2(at²)........................ Equation 1⇒ Where:
x = Final position of the objectx₀ = Starting positionv₀ = Starting velocityt = timea = accelerationFrom the question,
⇒ Given:
x₀ = 4.5 m/st = 2 sx₀ = 2ma = 0 m/s²⇒ Substitute these values into equation 1
x = 2+(4.5×2)+1/2(0²×2)x = 2+9+0x = 11 mHence, The final position of the object after 2 s is 11 m
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The fundamental of a closed organ pipe is 259.6 Hz. The second harmonic of an open organ pipe has the same frequency. What is the length of the closed pipe
Answer:
A closed organ pipe is λ/4 (node-antinode) long.
λ = speed / frequency = 331.5 / 259.6 = 1.28 m
λ/4 = .319 m length of closed pipe
An open pipe has a fundamental wavelength of A-N-A or λ/2
The second harmonic would be A-N-A-N-A or λ = 1.28 m for the second harmonic 331.5 / 259.6 = 1.28 (the fundamental would be 331.5 / .628
How long does it take the second hand of a clock to move through 3.57 rad?
Answer in units of s.
Answer:
Its angular velocity is
π
30
radians per second (about 0.105 radians/s.
Explanation:
Explanation:
Think of it like this:
There are
2
π
radians in one complete rotation, and that takes the second hand 60 seconds to complete.
So, the rate of rotation (the angular velocity) is
2
π
/60 = pi/30 radians per second# which is about 0.105 radians per second.
To solve the problem we will first calculate the time taken by the second arm to travel 1 rad.
Time taken by the second arm to travel 1 rad
We know that the arm of the second travels 2π radians in 60 seconds. therefore, the time is taken to travel 1 rad by the second arm,
[tex]1\rm\ rad = \dfrac{60}{2\pi}[/tex]
Time taken by the second arm to travel 3.57 radTime taken by the second arm to travel 3.57 rad
[tex]=3.57\times \dfrac{60}{2\pi}\\\\= 34.09\rm\ sec[/tex]
Hence, the time taken by the second arm to travel 3.57 rad is 34.09 sec.
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Where do hyperbolic comets originate?
A. the Oort cloud
B. the asteroid belt
C. the Kuiper belt
D. interstellar space
Answer:A.the Oort cloud
Explanation:
Answer:
try answer A..the Ootor cloud
Need help asap
A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3]
Answer:
Explanation:
Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?
Obviously the sea is salt water so we can ignore ρ = 1000 kg/m³
1025 kg/m³(d m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa
d = 9.9450535...
d = 10 meters
That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa
so the robot would be a hair above its pressure limit before it even got in the water.
a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed of the car is
Answer:
The new speed of the car is 30 m/s.
Hope you could get an idea from here.
Doubt clarification - use comment section.
Blaine and her sister are identical twins riding roller coasters at Kinetic Kars. They each ride the
roller coaster on their own once. Next time, they ride the roller coaster together. On which ride
do you think they have the most kinetic energy? Explain your answer using information from class
activities.
Suppose a bulldozer is being hauled at 50 km/h to a construction site on the back of a flatbed truck. From which reference point would the bulldozer not appear to be moving?
Answer:
from someone watching the bulldozer from the cab of the truck or from elsewhere on the flatbed.
Explanation:
Find the magnitude of the potential difference between two points located 1.2 m apart in a uniform 700 N/C electric field, if a line between the points is parallel to the field. Express your answer in volts.
The magnitude of the potential difference between the two points is 840 V
To solve the problem above, we need to use the formula of Potential difference as related to distance and electric field.
Potential difference: This can be defined as the work required to move a unit charge from a point to another in an electric field.
⇒ Formula:
V = E×d.................. Equation 1⇒ Where:
V = Potential differenceE = Electric Fieldd = DistanceFrom the question,
⇒ Given:
E = 700 N/Cd = 1.2 m⇒ Substitute these values into equation 1
V = 700(1.2)V = 840 VHence, The magnitude of the potential difference between the two points is 840 V
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if a current of 2.0 A is flowing from point a to point b, the potential difference between Vb- Va (in V) is:
a. 6
b. 8
c. -6
d. -8
e. 22
Answer:
[tex]\huge\color{skyblue}\boxed{\colorbox{black}{Answer ☘}}[/tex]
[tex]total \: resistance \: ( R_{t} ) = (3 + 1)ohm \\ (since \: the \: resistors \: are \: connected \\ in \: series)[/tex]
[tex]current \: flowing \: through \: circuit(I) = 2A \\ \\ now... \: by \: ohms \: law \\ V = IR\\ V = (2)(4) \\ = > 8v[/tex]
therefore , option (b) is correct!!hope helpful~
You hang a light in front of your house using an
elaborate system to keep the 12-kg object in static
equilibrium (Figure 1). What are the magnitudes of the
forces that the ropes must exert on the knot connecting
the three ropes if 02 = 639 and 03 = 45° ?
The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N F₂ = 89.21 N F₃ = 57.28 NGiven data :
Mass ( M ) = 12 kg
∅₂ = 63°
∅₃ = 45°
Determine the magnitudes of the forces exerted by the ropes on the connecting knota) Force exerted by the first rope = weight of rope
∴ F₁ = mg
= 12 * 9.81 ≈ 118 kg
b) Force exerted by the second rope
applying equilibrium condition of force in the vertical direction
F₂ sin∅₂ + F₃ sin∅₃ - mg = 0 ---- ( 1 )
where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 ) applying equilibrium condition of force in the horizontal direction
Back to equation ( 1 )
F₂ = [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]
= [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]
= 89.21 N
C ) Force exerted by the third rope
Applying equation ( 2 )
F₃ = ( F₂ cos∅₂ / cos∅₃ )
= ( 89.21 * cos 63 / cos 45 )
= 57.28 N
Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N
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What can happen when untreated sewage water contaminates water?
Answer:
Health Effects ↯Life-threatening human pathogens carried by sewage include cholera, typhoid and dysentery. Other diseases resulting from sewage contamination of water include schistosomiasis, hepatitis A, intestinal nematode infections, and numerous others.
Explanation:
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____
What net force acting on a 6 kg car produces an acceleration of 6 m/s2?
Explanation:
Mass (m) = 6 kgAcceleration (a) = 6 m/s²Force (F) = ?We know that,
• F = ma→ F = (6 × 6) N
→ F = 36 N
Therefore, the net force is 36N.
Which statement is true for a series circuit
Answer: they have one path to flow
Explanation: share the same current
PLEASE HELP ASP WILL GIVE 50 POINT AND BRAINLIEST!!!!!!!!!!!!!
A. In the Bohr model of the hydrogen atom, the speed of the electron is approximately
2.16 × 10⁶ m/s. Find the central force acting on the electron as it revolves in a circular orbit of radius 5.17 × 10⁻¹¹ m.
Answer in units of N.
B. Find the centripetal acceleration of the electron
Explanation:
A. The centripetal force experienced by an electron as it goes around a hydrogen nucleus is given by
[tex]F_c = m_e\dfrac{v^2}{r}[/tex]
where [tex]m_e = \text{electron\:mass} = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]r = 5.17×10^{-11}\:\text{m}[/tex] = orbital radius
[tex]v = 2.16×10^6\;\text{m/s}[/tex] = orbital velocity
so the centripetal force is
[tex]F_c = (9.11×10^{-31}\:\text{kg})\dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=8.22×10^{-8}\:\text{N}[/tex]
B. The electron's centripetal acceleration is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
Using the values from (A), we get
[tex]a_c = \dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=9.02×10^{22}\:\text{m/s}^2[/tex]
What is the half-life of the imaginary element Lokium? Show your work
Answer:What is the half life of the element Lokium?
The half-live of the element Lokium is 4.
Explanation:
sorry if im wrong, have good day
Answer:
The half-life of the element Lokium is 4
What is half-life?
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
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a car moving at 5 m/s accelerates at a rate of 10 m/s2 for 25 seconds. How far does it move during this time?
Answer:
t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2
Explanation:
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Describe what happens when iron and oxygen combine. Can the change be reversed?
Answer:
iron combines with oxygen to produce rust, which is the compound named iron oxide.
Explanation:
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
When Elements neutral atom contains 5 neutrons 4 electrons and 4 protons
Answer:
What is the atomic number of an atom that has 5 neutrons and 4 electrons? A neutral atom will have the same number of electrons is does protons. Since it has 4 protons, it must have an atomic number of 4. (That makes it beryllium.)
Explanation:
mark me brainliest please and thank you Ma'aM/Sir
Answer:
4 protons, 5 neutrons, and 4 electrons are there in an atom of beryllium.
Two Blocks are connected by a massless rope over a massless,
frictionless pulley, as shown in the figure. The mass of block 2
is m2 = 10.1 kg, and the coefficient of kinetic friction
between block 2 and the incline is Mk = 0.200. The angle 0 of
the incline is 27.5º. If block 2 is moving up the incline at
constant speed, what is the mass mi of block 1?
The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.
What is tension force?The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.
Given that,
Mass of block 1=? kg
The coefficient of the kinetic friction,μ=0.200
Now consider the weight component in the uphill direction.The weight is found as;
[tex]\rm W=m_1gsin \theta[/tex]
The force is balanced in the vertical direction as;
[tex]\rm T=F_f-W[/tex]
When the force of friction is;
[tex]\rm F_F=\mu_k N[/tex]
[tex]\rm F_f=(m_1 gcos \theta)[/tex]
Substitute the value in the vertical balanced equation;
[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]
[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]
[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]
Hence the mass of block 1 will be 1.99 kg.
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Hey guys can you help me with this problem it's giving me a headache. "If 2500 kg roller coaster car begins its descent from 500m above above the first hill. What is its kinetic energy at the bottom of the hill."
The kinetic energy of the roller coaster at the bottom of the hill is 12250000 J
To answer this question, we must understand that the energy of a system is always conserved as explained by the law of conservation of energy.
The kinetic energy of the roller coaster will be equivalent to the potential energy of the roller coaster.
The kinetic energy of the roller coaster can be obtained as follow:
Mass (m) = 2500 KgHeight (h) = 500 mAcceleration due to gravity (g) = 9.8 m/s²Kinetic energy (KE) =?Kinetic energy = Potential energy
KE = mgh
KE = 2500 × 9.8 × 500
KE = 12250000 J
Therefore, the kinetic energy of the roller coaster is 12250000 J
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9. P waves move faster than S waves A. True B. False
Convert the decimal number 61078 to binary by using sum-of-weights method
Answer:
1110111010010110
Explanation:
I am not able to upload the working out using the sum of weights method sorry
Sophie applies a 50 n force to push a box 2 meter across the floor calculate the smount of work done in the box
A 2N and an 6N force pull on an object to the right and a 4N force pulls on the object to the left. If the object has a mass of 0.25 kg what is its acceleration?
Answer:
[tex]16m/s^{2}[/tex]
Explanation:
Please ask if you have more questions!