Answer:
C
Explanation:
only if there is a net force of zero, the body will not move
some people may say B but that is wrong because maybe one force is greater than the other so the object would still move even though the forces are in opposite directions and parallel
the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. what is the radius of the moon?
Given that,
The earth's moon has a gravitational field strength of about 1.6 n/kg
Mass of Moon, \(M=7.35\times 10^{22}\ kg\)
To find,
The radius of the Moon.
Solution,
The formula for the acceleration due to gravity is given by :
\(g=\dfrac{GM}{r^2}\)
r is radius of the Moon
\(r=\sqrt{\dfrac{GM}{g}} \\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.6}} \\\\r=1750437.44\ m\\\\r=1.75\times 10^6\ m\)
So, the radius of the Moon is \(1.75\times 10^6\ m\).
A white pool ball of mass 0.5 kg moving at 10 m/s collides with a 0.4
kg red pool ball initially at rest. After the collision, the white pool ball
continues in the same direction with a velocity of 1.1 m/s. Neglecting
friction, the velocity of the red ball after the collision is m/s.
The velocity of the red ball after the collision is m/s is 11. 125 m/s
The conservation of momentum states that, inside some hassle domain, the quantity of momentum stays regular; momentum is neither created nor destroyed, but handiest modified thru the action of forces as defined via Newton's legal guidelines of motion.
Calculation:-
Given,
M = 0.5-kg
V i = 10 m/s
M₂ = 0.4 kg
V i = 0
V f = 1.1 m/s
V f = ?
M₁V₁ + M₂V₂ = V f M₁ + V f M₂
= 0.5 × 10 + 0.4 ×0 = 1.1 × 0.5 + V f × 0.4
= 5 = 0.55 + 0.4 V f
V f = 5 - 0.55/0.4
= 4.45 / 0.4
= 11. 125 m/s
The law of conservation of momentum states that in an isolated machine the whole momentum of two or extra our bodies performing upon each different remains regular unless an outside pressure is applied. therefore, momentum can neither be created nor destroyed.
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The bolts on a car wheel require tightening to a torque of 4 N*m. If a 9 cm long wrench is used, what is the magnitude of the force required when the force is applied at 8 degrees to the wrench
Answer:
320N
Explanation:
The magnitude of the torque required is expressed using the formula;
T = Fr sin theta where;
F is the force
r is the radius = 9cm = 0.09m
theta is the angle of inclination = 8 degrees
Torque T = 4Nm
Substitute the given values and get F
4 = F(0.09)sin8
4 = 0.0125F
F = 4/0.0125
F = 320N
Hence the magnitude of the force required when the force is applied at 8 degrees to the wrench is 320N
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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represent 11 by 9 on a number line
A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later. Determine the distance the ball was thrown.
The distance the ball was thrown is 52.22m by A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later.
How to calculate distance?Every motion under constant acceleration is projectile motion
Angle above horizontal Ф = 35°, initial speed v1 = 25m/s , time 2.55s
Substituting value in the below equation
x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²
a= 0 as acceleration in horizontal direction is zero
x= 25*cos(35)*2.55
x=52.22 m
Projectile motion is a form of motion in which object influenced when it is launched into the gravitational force from the surface of Earth along a curved path.
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11.) If figure skater spins with a circular acceleration of 275 m/s^2 in a circle 5 poin with a radius of 0.55 meters, calculate the velocity of the skater."
Answer:
we don't care what ure saying thats absurd
Part E
Once you complete your outline, write a 500- to 750-word paper using word processing software. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.
A Voyage to Proxima Centauri
Proxima Centauri is the second closest star to our solar system, which makes scientists curious to know more about it. What do we know about Proxima Centauri? How did we discover this information? Is it possible for humans to travel to this star? In this task, you will research and write a 500- to 750 word paper that answers these questions. Follow these steps to complete your research and writing. This guide about the research process can help.
Estimated time to complete: 3 hours
Part A
The goal of your paper is to describe the history of the discovery and research of Proxima Centauri. Another goal is to find the possibilities of traveling to this star. Some questions that your paper should answer are:
When was Proxima Centauri discovered?
How was Proxima Centauri discovered?
How have scientists researched Proxima Centauri?
What technologies have they used? What types of data do these technologies collect?
Have spacecraft ever reached this star?
What are the limitations of sending spacecraft to Proxima Centauri?
What accommodations would humans need to travel to Proxima Centauri?
Proxima Centauri otherwise known as Alpha Centauri C, is the closest star to our solar system, located about 4.24 light-years away. It was discovered in 1915 by Robert Innes, the director of the Union Observatory in South Africa. Innes identified Proxima Centauri as a possible member of the Alpha Centauri system after noticing that the star had a similar proper motion to Alpha Centauri A and B.
Proxima Centauri was first observed through telescopes, which allowed scientists to study the star's spectra and estimate its distance from Earth. In the following decades, scientists continued to study Proxima Centauri through telescopes, including the Hubble Space Telescope, which provided high-resolution images of the star.
In 2016, the European Southern Observatory announced the discovery of an exoplanet orbiting Proxima Centauri, called Proxima Centauri b. This discovery was made using the radial velocity method, which measures the star's small wobbles caused by the gravitational pull of an orbiting planet.
What are the major challenges facing the discovery of Proxima Centauri?Despite the advances in our understanding of Proxima Centauri, no spacecraft have ever reached this star. The distance between Proxima Centauri and Earth is so vast that it would take tens of thousands of years to reach the star using current propulsion technology.
One potential solution to this problem is the use of interstellar travel, which would allow humans to travel to Proxima Centauri within a human lifetime. There are several proposed methods for interstellar travel, including using a starship propelled by fusion engines or a massive light sail pushed by a beam of lasers. However, these technologies are still in the theoretical stage and have not yet been developed.
There are also many other challenges that must be overcome in order for humans to travel to Proxima Centauri. For example, humans would need to find a way to protect themselves from the high levels of radiation that they would be exposed to during the journey. They would also need to find a way to provide enough food, water, and other resources to sustain themselves for the duration of the trip.
Overall, while it is theoretically possible for humans to travel to Proxima Centauri, it would be a massive undertaking that would require significant technological advancements and the development of new solutions to many challenges.
Sources of Works Cited include:
"Proxima Centauri." Wikipedia, Wikimedia Foundation, 5 Jan. 2021,
"Proxima Centauri b." Wikipedia, Wikimedia Foundation, 20 Nov. 2020,
"Interstellar Travel." Wikipedia, Wikimedia Foundation, 4 Jan. 2021,
Therefore, the correct answer is as given above
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Mrs Turner has a scale drawing of her apartment.IN the scale drawing the lenght of her apartment is 6.25 inches and the width is 2.75 onches. If Mrs. Turner used the scale factor 1 inc/18ft, what would the area of the apartment be?
The area of Mrs. Turner's apartment is approximately 0.05304 square feet.
To find the area of Mrs. Turner's apartment, we need to convert the measurements from the scale drawing to the actual measurements in feet, using the given scale factor of 1 inch/18 feet.
Length of apartment in feet = 6.25 inches × (1 foot/18 inches) = 0.34722 feet
Width of apartment in feet = 2.75 inches × (1 foot/18 inches) = 0.15278 feet
Now, we can calculate the area of the apartment in square feet by multiplying the length and width:
Area of apartment in square feet = Length × Width = 0.34722 feet × 0.15278 feet = 0.05304 square feet
Therefore, the area of Mrs. Turner's apartment is approximately 0.05304 square feet.
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S= 12 m/s D= ?t= 3 s What distance did the object traveling at 12m/s go in 10 points
3s? *
4 m
3 m
36 m
Answer:
distance travelled= 12 x 3=36 m
1. A bullet is fired into a tree trunk (Figure 1.12), striking it with an ini- tial velocity of 350 m/s. If the bullet penetrates the tree trunk to a depth of 8.0 cm and comes to rest, what is the acceleration of the bullet?
rtdgsehrtejffej0.345
A 50.0kg woman climbs a flight of stars 6.00 high in 15.0s. How much power dose she use?
First, calculate the weight
Weigth = mass x gravity = 50 kg x 9.8 m/s^2 = 490 Newton
Then, the work
Work = Weight x height = 490 N x 6m
finally, calculate the power:
Power = Work / time = (490N x 6m) / 15s
Power = 196 newton * m /s
Power = 190 watts
Chemical A and Chemical B react in an exothermic reaction. What can be known about what will happen when Chemical A and Chemical B are mixed together?
The old substance will need more energy to break its chemical bonds than the new substance will release.
More energy will be released from the old substance than the new substance will release to form its chemical bonds.
The color will change as a result of the reaction.
The substance will bubble as a result of the reaction.
In an exothermic reaction, more energy will be liberated from the old substance than the new substance will liberate to form its chemical bonds. Therefore, option B is correct.
What is an exothermic process?An exothermic process can be described as a thermodynamic reaction that emits energy from the system to the surrounding in the form of heat, light, or sound.
In chemical reactions, the bond energy is converted into thermal energy. In exothermic reactions, the reaction takes place in the form of the kinetic energy of molecules. The electronic transition of electrons from one energy level to another leads light to be released.
The energy of the reactants is more than the energy of the products in the exothermic reaction. The energy liberated by reactants will be more than the energy required for the new bonds to be formed in the products.
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A wave has an amplitude of 0.0800 m
and is moving 7.33 m/s. One oscillator
in the wave takes 0.230 s to go from
one crest to the next crest. Find the
wavelength of the wave.
(Unit = m)
If a wave has an amplitude of 0.0800 m and is moving 7.33 m/s. The
wavelength of the wave is 1.69m.
What is the wavelength?The wavelength of a wave can be determined using the equation:
Wavelength = velocity / frequency
To determine the frequency we need to calculate the reciprocal of the time it takes for one complete oscillation.
frequency = 1 / time
frequency = 1 / 0.230
frequency ≈ 4.35 Hz
Substitute the values into the wavelength equation:
wavelength = velocity / frequency
wavelength = 7.33 / 4.35
wavelength ≈ 1.69m
Therefore the wavelength of the wave is approximately 1.69 meters.
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A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (3.0 - 2.0) m/s2. At the instant the particle moves back across the x axis (y = 0), what is the value of its x coordinate?
When the particle reaches its maximum positive x coordinate, the displacement is 11.62m.
What is acceleration ?The rate at which speed of velocity change over time is referred to as acceleration. When a point and object speeds or decelerates, it is moving straight ahead. Motion on a circular accelerates even while the speed is constant so because orientation is constantly changing.
Briefing:The particle's displacement is computed as follows;
v² = u² + 2as
where;
a is acceleration
a = √(3² - 2²)
= 2.23 m/s²
0 = u² + 2as
s = -u²/2a
s = -(7.2²)/(2 x 2.23)
s = -11.62 m
When the particle reaches its maximum positive x coordinate, the displacement is 11.62m.
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audible limit of the sound wave
Answer:
20 hz - 20khz is the audible sound wave for humans
Almost 18Hz is an audible limit of the sound wave.
Need help ASAP plsss
Write 5-6 sentences about
:how do parachutes work ??
A ripcord system pulls a closing pin, which releases a spring-loaded pilot chute, and opens the container; from which the parachute comes out.
If a skydiver is unable to open her/his parachute, an automatic activation device will automatically open the parachute.
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines which change the shape of the wing, cause it to turn, or to increase or decrease its rate of descent.
A cart moving across a level surface accelerates
uniformly at 1.0 meter per second² for 2.0
seconds. What additional information is
required to determine the distance traveled by
the cart during this 2.0-second interval?
The initial velocity of the cart.
Newton, There can be a mass is four.080, So acceleration might be equal to 2.50 m in step with cent within the rectangular. Initial is that amount that relies upon total mass. The greater mass the more inertia. So Mass is 2000 kg and acceleration is 3 ms square. So this offers us an internet pressure identical to 6000 newtons or 6.0 and 210 to the power
In case you roll a ball, it initially will keep rolling except friction or something else stops it by means of pressure. you could also think about the way that your body maintains transferring ahead when you hit the brake on your bike. Translational Inertia = ma, in which m is the mass, and a is the acceleration of the object. Calculate the rotational inertia or the instant of inertia velocity by way of multiplying the mass of the object with a square of the gap between the item and the axis, the radius of rotation.
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Lab Report
Light
It’s time to complete your Lab Report. Save the lab to your computer with the correct unit number, lab name, and your name at the end of the file name (e.g., U4_ Lab_Light_Alice_Jones.doc).
Introduction
1. What was the purpose of the experiment?
Type your answer here:
2. What were the independent, dependent, and control variables in your investigation? Describe the variables for each part of the experiment.
Type your answer here:
Experimental Methods
1. What tools did you use to collect your data?
Type your answer here:
2. Describe the procedure that you followed to collect your data.
Type your answer here:
Data and Observations
1. Record your observations in the data tables.
Type your answer here:
Table 1. Angles of Incidence and Angles of Reflection for a Beam of Light Reflected off a Mirror
Angle of incidence (°) Angle of reflection (°)
Table 2. Angles of Incidence and Angles of Refraction for a Beam of Light Refracted through a Glass Lens
Angle of incidence (°) Angle of refraction (°)
Conclusions
1. Draw a diagram (develop a model) that shows what happens to light when it reflects off a mirror. Look for patterns in your data to help you develop your model. Hint: Is the angle of reflection always greater than, less than, or equal to the angle of incidence? Include labels.
Type your answer here:
2. Use your model to predict the angle of reflection of a beam of light reflected off a mirror if the angle of incidence of the beam of light is 40 degrees.
Type your answer here:
3. Draw a diagram (develop a model) that shows what happens to light when it is refracted through a glass lens. Look for patterns in your data to help you develop your model. Hint: Is the angle of refraction always greater than, less than, or equal to the angle of incidence? Include labels.
Type your answer here:
Answer:
Explanation:
the answer is b
if something is frictionless does it have thermal energy
Answer:
No
Explanation:
because there is no pressure
For the system of Problem3.18, determine the amount of energy transfer by heat, in kJ per kg of refrigerant.[Problem3.18A closed, rigid tank contains a two‐phase liquid–vapor mixture of Refrigerant 22 initially at−20°Cwith a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. Determine the final temperature, in°C. If the final state is in the superheated vapor region, at what temperature, in°C, does the tank contain only saturated vapor?]
Answer:
35.625° C
Explanation:
See attachment
a) Final temperature is 35.625° C
b) 0°
If we were to make water (H20) based on the Key Provided above,what would the water molecule look like
Answer:
Individual H2O molecules are V-shaped, consisting of two hydrogen atoms (depicted in white) attached to the sides of a single oxygen atom (depicted in red). Neighboring H2O molecules interact transiently by way of hydrogen bonds (depicted as blue and white ovals).
A flat sheet of paper of area 0.450 m2 is oriented so that the normal to the sheet is at an angle of 600 to a uniform electric field of magnitude 18 N C-1. What is the magnitude of the electric flux through the sheet? A. 3.22 N m2 C-1 B. 21.42 N m2 C-1 C. 5.04 N m2 C-1 D. 11.72 N m2 C-1 E. 4.05 N m2 C
The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).
The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.
In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²
(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:
Electric flux = Electric field strength × Area
Electric flux = 18 N C⁻¹ × 0.450 m²
Electric flux = 8.1 N m² C⁻¹
In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.
The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).
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A 80kg stone falls from the top of the 360 meter cliff. Neglecting friction, how fast will the stone be moving just before it hits the ground?
The stone will be moving at a speed of approximately 84.4 meters per second just before it hits the ground, neglecting friction.
To find how fast will the stone be moving just before it hits the ground?This problem can be solved using the laws of kinematics and conservation of energy. The potential energy of the stone at the top of the cliff is converted to kinetic energy as it falls. We can equate the potential energy at the top of the cliff to the kinetic energy just before hitting the ground.
Potential energy = mgh,
Where
m is the mass of the stone g is the acceleration due to gravity (9.8 m/s^2) h is the height of the cliff (360 meters)Kinetic energy = (1/2)mv^2,
Where
v is the velocity of the stone just before hitting the ground.Equating these two expressions and solving for v, we get:
mgh = (1/2)mv^2
v^2 = 2gh
v = sqrt(2gh)
Plugging in the given values, we get:
v = sqrt(2 x 9.8 m/s^2 x 360 m) = 84.4 m/s
Therefore, the stone will be moving at a speed of approximately 84.4 meters per second just before it hits the ground, neglecting friction.
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A 0.1 kg tennis ball is brought from a speed of 15 m/s to a speed of 45 m/s when hit by a tennis racket. If this done over 0.05 seconds, what is the power generated by the tennis racket
A 750 kg race car accelerates to the right. The engine applies a force of
+7000 N to the car. The total friction on the car in the opposite
direction is -200 N.
Find the net force on the car then use 'Fnet = m a' to find the
acceleration.
O a
b
Oc
Od
12.4 m/s2
10.5 m/s2
8.0 m/s2
9.1 m/s2
in a typical cop movie we see the hero pulling a gun firing that gun straight up into the air and shouting
It is not recommended to fire a gun straight up into the air.
When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.
Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.
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A high-speed photograph of a club hitting a golf ball is shown in the figure below. The club was in contact with a ball, initially at rest, for about 0.0052 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 1.6 102 ft/s, find the average force exerted on the ball by the club.
Given:
The time duration for which the club was in contact with the ball, t=0.0052 s
The mass of the ball, m=55 g=55×10⁻³ kg
The initial speed of the ball, u=0 m/s
The final speed with which the ball leaves the club, v=1.6×10² ft/s=48.77 m/s
To find:
The force exerted on the ball, F.
Explanation:
The acceleration of an object is the time rate of change of velocity.
Thus the acceleration of the ball is given by,
\(a=\frac{v-u}{t}\)On substituting the known values,
\(\begin{gathered} a=\frac{48.77-0}{0.0052} \\ =9378.8\text{ m/s}^2 \end{gathered}\)From Newton's second law of motion,
\(F=ma\)On substituting the known values,
\(\begin{gathered} F=55\times10^{-3}\times9378.8 \\ =515.8\text{ N} \end{gathered}\)Final answer:
Thus the average force exerted on the ball is 515.8 N
What is the mass of a car possessing 15,000 kg*m/s of momentum going a speed of 35 m/s?
A.525000 kg
B.428.6 kg
C.14965 kg
D.0.0023 kg
Answer: 30,000 kg m/
Explanation: Therefore p=1500*20 = 30,000 kg m/s.
The 1kg rock is tied to a string and swung in a circular path as shown. The 1 meter string is tied to a post, and during the motion, the string has a 30 angle with the post. The rock makes 100 rounds in 1 minute. The centripetal force on the rock is
Answer: 54.8
Explanation:
The centripetal force on the rock moving along the string is 104.66 N.
The given parameters;
the mass of the rock, m = 1 kgangle of inclination of the string, θ = 30⁰angular velocity of the rock, ω = 100 rev/minlength of the string, r = 1 mThe centripetal acceleration of the rock is calculated as follows;
\(\omega _f^2 = \omega _i^2 + 2\alpha (\theta)\\\\(100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} )^2 = 0 + 2\alpha (30 ^0 \times \frac{\pi \ rad}{180^0} )\\\\109.69 = 1.048\alpha \\\\\alpha = \frac{109.69}{1.048} \\\\\alpha =104.66 \ rad/s^2\)
\(a_c = \alpha \times r\\\\a_c = 104.66 \times 1 = 104.66 \ m/s^2\)
The centripetal force on the rock is calculated as follows;
\(F_c = ma_c\)
\(F_c = 104.66 \times 1\\\\F_c = 104.66 \ N\)
Thus, the centripetal force on the rock moving along the string is 104.66 N.
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