Answer:
I don't know how to do it the subject
The centripetal acceleration of the car when there is the radius and the speed is given so it should be considered as the option b. 36 [tex]m/s^2[/tex].
Calculation of the centripetal acceleration of the car:Since
The radius is 25 cm and the speed should be 30 m/s
Now the following formula should be used
Acceleration = speed^2/ radius
= 30^2 / 25
= 900 / 25
= 36 [tex]m/s^2[/tex]
Therefore, The centripetal acceleration of the car should be considered as the option b. 36 [tex]m/s^2[/tex].
Learn more about speed here: https://brainly.com/question/20361733?referrer=searchResults
A mom pushes her 19.3 kg daughter on the swing. If she gives her an initial velocity of 7.5 m/s at the bottom of the swing and the swing sits 0.6 m above the ground at it's lowest point, what height does she reach above the ground?
Answer:
3.17333333333? I hope I get it right
Explanation:
..................hello
Somebody, please help me with these by April 6 I'm about to fail my class
1. Describe specific heat capacity and its use in calorimetry.
2. Define latent heat and how it is different than specific heat capacity.
3. Describe how a phase diagram changes when changing from a solid to a liquid.
4. Describe how work is done is related to a change in the volume of a fluid.
—Continue on the next page—
5. What is the relationship of change in internal energy, work, and heat? Explain the significance in terms of energy.
6. As succinctly as possible, explain why it is impossible to have an engine that is 100% efficient.
7. Describe the application of Archimedes principle in everyday terms.
Practice
8. 50 grams of a substance increases its temperature by 10 degrees when 100 J of heat is added. What is the specific heat capacity of the substance?
9. Ten kilograms of a substance has a latent heat of 1000 J/kg. How much energy does it take to change the phase of this substance?
—Continue on the next page—
10. The work done by a piston is measured to be 1000 J. If the pressure is a constant 1000 Pa, what is the change in volume of the piston?
11. Is the process described in #10 isovolumetric? Explain.
12. Is it possible for a substance to change the temperature in an isovolumetric and adiabatic process? Explain.
13. A substance is heated with 1000 J and does 700 J of work on the atmosphere. What is the change in the internal energy of the substance?
14. An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency of this engine?
—Continue on the next page—
15. If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant force?
16. A 10 kg ball of volume = 0.005 m3 is set on a lake. Describe what will happen to the ball.
Application
A group of physics students heats a vial of water, which has a balloon at the top that traps the water vapor and air.
17. Describe how much the internal energy is changed if the students are adding heat to the water via a Bunsen burner. No need to calculate anything.
18. If 50 grams of water start to boil, then how much heat must be added to completely boil off the water? The latent heat of vaporization is 2,260,000 J/kg
—Continue on the next page—
19. Assume this process is 10% efficient, how much work is done by the gas expanding into the atmosphere?
20. Assuming the atmospheric pressure to be constant at 101,325 Pa, by what amount does the volume of the balloon change?
Answer:
im sorry i would help but thats too much
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
Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller.
a. True
b. False
Help please. Question about a potential energy.
A wire carries a current of 4.2 A at what distance from the wire does the magnetic field have a magnitude of 1.3×10^ -5 t
Answer:
the distance is 6.46 cm.
Explanation:
Given
current in the wire, I = 4.2 A
magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T
The distance from the wire is determined by using Biot-Savart Law;
[tex]B = \frac{\mu_o I}{2\pi r} \\\\r = \frac{\mu_o I}{2\pi B}[/tex]
Where;
r is the distance from the wire where the magnetic field is experienced
[tex]r = \frac{\mu_o I}{2\pi B}\\\\r = \frac{4\pi \times 10^{-7} \times 4.2 }{2\pi \times 1.3 \times 10^{-5}}\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm[/tex]
Therefore, the distance is 6.46 cm.
6. The rate at which velocity changes is called
O speed
O direction
O acceleration
O displacement
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
Please help I will mark you brainliest
I believe the answer is a
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 spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.62c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The veolcity of Enterprise 2 relative to Enterprise 1 is 0.30c. What is the velocity of Enterprise 2
Answer:
The answer is "0.92 c"
Explanation:
[tex]v_1\ (earth) = 0.62 \ c \\\\v_2\ ( enterprise ) = -0.30[/tex]
so,
[tex]v_2 \ (earth) = 0.62 \ c - (-0.30 \ c) \\\\[/tex]
[tex]= 0.62 \ c +0.30 \ c\\\\= 0.92 \ c[/tex]
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)
Force of a Baseball Swing. A baseball has mass 0.153 kg . Part A If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 50.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. Express your answer to three significant figures and include the appropriate units. P
Answer: 14.5 kg.m/s
Explanation:
Given
mass of baseball is [tex]m=0.153\ kg[/tex]
The initial speed of the ball is [tex]u=-44.5\ m/s[/tex]
the final speed of the ball is [tex]v=50.5\ m/s[/tex]
Impulse is given as a change in the momentum
[tex]\vec{J}=\Delta \vec{P}[/tex]
[tex]J=m(v-u)\\J=0.153(50.5-(44.5))\\J=0.153\times 95=14.535\ kg.m/s[/tex]
Change in momentum up to 3 significant figures is 14.5 kg.m/s
Impulse applied by a bat is also the same as the change in momentum
A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?
A student using a stopwatch finds that the time for 10 complete orbits of a ball on the end of a string is 25 seconds. The period of the orbiting ball is
Answer:
T = 2.5 s
Explanation:
Given that,
Number of complete orbits = 10
Time, t = 25 seconds
We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.
So,
[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]
So, the period of the orbiting ball is equal to 2.5 seconds.
When you cool a gas, how does this affect the de Broglie wavelength of the gas atoms? When you cool a gas, how does this affect the de Broglie wavelength of the gas atoms? Being cooled, the gas atoms slow down so that their de Broglie wavelength will increase. Being cooled, the gas atoms slow down so that their de Broglie wavelength will decrease. The de Broglie wavelength will remain the same because it does not depend on temperature.
Answer:
The de Broglie wavelength will remain the same because it does not depend on temperature.
Explanation:
de Broglie wavelength of a particle is independent of the temperature and hence the properties of emitted particle such as photoelectric effect, radioactive radiation etc. does not depend on the temperature.
Also, until unless the kinetic energy of a moving particle is not driven by the
thermal energy, the de Broglie wavelength is independent of the temperature
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a) v = 1.1 m/s
b) A = 0.315 m
c) v = 1.1 m/s A= 0.15 m
Explanation:
a)
In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:[tex]v = \lambda * f (1)[/tex]
If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:⇒ T = 2.60 s * 2 = 5.20 s (2)Since we have now T, we can find the frequency f as follows:[tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]
Replacing f and λ in (1) we get:[tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]
b)
The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:[tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]
c)
Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:v = 1.10 m/s (6)
Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:[tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]
It turns out that the depth in the ocean to which airborne electromagnetic signals can be detected grows with the wavelength. Therefore, the military got the idea of using very long wavelengths corresponding to about 30 Hz to communicate with submarines throughout the world. If we want to have an antenna that is about one-half wavelength long, how long would that be
Wavelength = speed / frequency.
Wavelength = 3x10^8 m/s / 30 hz
Wavelength = 10 million meters
1/2 wavelength = 5 million meters
(that's about 3,100 miles)
I'm pretty sure the frequency is wrong in the question.
I think it's actually 30 kHz, not 30 Hz.
That makes the antenna about 3.1 miles long.
Find the wavelength of light which is capable of ionizing a hydrogen atom?
Answer:
The correct answer is - 91.4 nm
Explanation:
According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:
(h×c)/λ=13.6eV
here,
h - Planck constant
c - the speed of light
λ - wavelength
Placing the value in the formula for the wavelength
(6.626×10^−34J.s × 3×10^8 m/s)/λ = 13.6 ×1.6 × 10^−19 J
λ≈91.4nm
Thus, the correct answer would be = 91.4 nm
Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as proton two. The magnitude of the magnetic field in which proton one is immersed is twice the magnitude of the magnetic field in which proton two is immersed. The radius of the circle around which proton one moves is r. What is the radius of the circle around which proton two moves
Answer:
r₂ = 4 r
Explanation:
For this exercise let's use Newton's second law with the magnetic force
F = q v x B
bold letters indicate vectors, the magnitude of this expression is
F = q v B sin θ
in this case we assume that the angle is 90º between the speed and the magnetic field.
If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal
a = v² / r
let's use Newton's second law
F = ma
q v B = m v² / r
r = [tex]\frac{qB}{mv}[/tex]
Let's apply this expression to our case.
Proton 1
r = \frac{qB_1}{mv_1}
Proton 2
r₂ = [tex]\frac{q \ B_2}{m \ v_2}[/tex]
in the exercise indicate some relationships between the two protons
* v₁ = 2 v₂
v₂ = v₁ / 2
* B₂ = 2B₁
we substitute
r₂ = [tex]\frac{q \ 2B_1}{m \ \frac{v_1}{2} }[/tex]
r₂ = 4 [tex]\frac{qB_1}{mv_1}[/tex]
r₂ = 4 r
volcano has both useful and harmful effects give reason
Answer:
harmful effects
1. that will cause air pollution
2. that will destroy our earth
Answer:
useful effects of volcano are :-
it makes soil fertile it provides valuable nutrients for the soilharmful effects of volcano are:-
it makes air polluted it destroy the environment .hope it is helpful to you ☺️
At which point is there the most potential energy? At which point is there the most kinetic energy?
A. Potential energy A; Kinetic energy B
B. Potential energy B; Kinetic energy D
C. Potential energy A; Kinetic energy D
D. Potential energy C; Kinetic energy D
Answer:
The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy
2. Which of the following objects has the greatest momentum?
A) a 145 grain baseball hit at 45 m/s
B) a 1200 kg car driving 25 m/s
C) a 2.0 *10-2 kg snail moving al 3.0 * 104 m/s
D) a 500 kg horse galloping at 12.0 m/s
Answer:
B) a 1200 kg car driving 25 m/s
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
This ultimately implies that, the mass of an object or body is directly proportional to its momentum. Thus, the higher the mass of an object or body, the greater would be its momentum and vice-versa.
By mere inspection of the data given, we can see that the object with the greatest amount of mass and velocity is the car weighing 1200 kilograms and moving at 25 meters per seconds.
Substituting into the formula, we have;
[tex] Momentum = 1200 * 25 [/tex]
Momentum = 30,000 Kgm/s
A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet
Answer:
[tex]291.67\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of bullet = 4 g
[tex]m_2[/tex] = Mass of block = 1.3 kg
[tex]\mu[/tex] = Coefficient of friction = 0.17
[tex]s[/tex] = Displacement of block = 0.24 m
[tex]v_1[/tex] = Velocity of bullet
[tex]v[/tex] = Velocity of combined mass
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]
As the momentum is conserved in the system we have
[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]
The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].
An artificial satellite circling the Earth completes each orbit in 126 minutes. (a) Find the altitude of the satellite.
Answer:
Explanation:
Time period of rotation
T = 2πR/ V where R is radius of orbit and V is orbital velocity
Orbital velocity V = √ ( GM/R ) , m is mass of the earth .
T = 2πR √R / GM
T² = 4π²R³ / GM
Putting the values
( 126 x 60 )² = 4 x 3.14² x R³ / 6.67 x 10⁻¹¹ x 5.97 x 10²⁴
57.15 x 10⁶ = 39.44 x R³ / 39.82 x 10¹³
R³ = 577 X 10¹⁸
R = 8.325 x 10⁶ m
= 8325 km
Radius of earth = 6400 km
height of satellite = 8325- 6400 = 1925 km .
Blocks A (mass 5.00 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line.
(a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) Find the velocity of each block after they have moved apart.
Answer: i believe its B Find the velocity of each block after they have moved apart sorry
Explanation: have a nice day buddy
Pls help ASAP
Imagine that Maritans launch a rocket toward the Earth at a great speed. While the
rocket is traveling toward us, it will appear
than it actually is.
O more blue
darker
larger
more red
Answer:
The rocket will appear larger than it actually is
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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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
The Solar System formed...
from the collision between two galaxies.
from the merging of two smaller stars.
directly from the leftover gas of the big bang.
from the gravitational collapse of an interstellar cloud of gas.
Answer:The answer is because of the gravity and the mass formed in the sun, the magnetic field reacts to it and leaves a fault on earth. wind goes by the earth and procides to be ok
Explanation:that is it