What is used to measure heat?

change in temperature
constant temperature
constant specific heat
change in specific heat

Answers

Answer 1

Answer:

change in temperature

Answer 2

Answer:

Change in temperature

Explanation:

did this on edge yesterday and that was the answer


Related Questions

Susan competes in luge competitions during the winter months. She rides solo on a small sled 3 inches off the ground down icy slopes, turning only by use of her feet and the shifting of her weight on the sled. During the initial stage of one downhill luge, Cynthia accelerated from rest at 5.9 m/s2 for 3.2 seconds. Calculate how far she traveled during this time.

I need this done and I need an explanation as to how I solve this problem step by step. YOU WILL NOT RECEIVE BRAINLIEST IF YOU DO NOT SHOW YOUR WORK .

Answers

The distance covered is 30.2 m

Explanation:

The motion of Cynthia is a uniformly accelerated motion (constant acceleration), so we can find the distance covered by using the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem,

u = 0 (since Cynthia started from rest)

t = 3.2 s

[tex]a=5.9 m/s^2[/tex] is her acceleration

Substituting,

[tex]s=0+\frac{1}{2}(5.9)(3.2)^2=30.2 m[/tex]

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Final answer:

Using the equation of motion S = ut + ½at2, we calculated that Susan traveled 30.216 meters during the initial stage of her downhill luge, where she accelerated at 5.9 m/s2 for 3.2 seconds.

Explanation:

To calculate how far Susan traveled during the initial stage of her downhill luge, where she accelerated from rest at 5.9 m/s2 for 3.2 seconds, we use the equation of motion for constant acceleration:


S = ut + ½at2

Here, S represents the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Since Susan starts from rest, her initial velocity u = 0. We plug in the acceleration a = 5.9 m/s2 and time t = 3.2 s into the equation:


S = (0)(3.2) + ½(5.9)(3.2)2

S = 0 + ½(5.9)(10.24)

S = 30.216 m

Therefore, Susan traveled 30.216 meters during the initial stage of her downhill luge.

What factors limit the accuracy of a potentiometer?

Answers

Answer:

Whats a potentiometer.

Explanation:

Some of the factors that can limit the accuracy of a potentiometer are internal temperature and Resistance change in the resistor

A potentiometer is a variable resistor that can be adjusted manually and has three terminals. A potentiometer is also called a pot

Further Explanation

Potentiometers have different shapes and they are used in different in different application. A potentiometer can be used to control the volume of a radio. There are several types of a potentiometer.

some of the common ones include

Single-turn pot: it is a single rotation of about 270 degreesMulti-turn pot: this type of pot is used where high precision and resolution are needed. Dual gang pot: this type of pot is used in audio volume control.Concentric pot: this type of pot is mostly used in older car radios, particularly in radios that have its volume and the tone control combinedServo pot: this type of pot is used where manual and automatic adjustment is needed.

However, some of the materials that can be used to construct potentiometers include

Metal firmCarbon compositionWirewoundCermet

As stated earlier, the two factors that can limit the accuracy of a potentiometer are:

Internal temperatureResistance change in the resistor

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potentiometeraccuracyfactorslimittypes of potentiometer

The driver of a car applies the brakes for 4.0 s.As a result an average force of 7500N is exerted in the car to slow it down.What is the change in momentum of the car.

Answers

Answer:

The change in momentum of the car is  30,000 kg m/s                  

Explanation:

Given,

The force exerted on the car to slow down, F = 7500 N

The time period of force, t = 4 s

The rate of change of momentum of the object is equal to the force acting on it.

Therefore,

                                    (mv - mu) / t = F

Where v and u are the final and initial velocity of the car. The change in momentum of the car,

                                    mv - mu = F x t

                                                   = 7500 x 4

                                                   = 30,000 kg m/s

Hence the change in momentum of the car is  30,000 kg m/s                  

Which would be the easiest way to move a heavy box up 4 feet? To lift it up. To slide it up a 8 foot ramp. To slide it up a 4 foot ramp. To slide it up a 2 foot ramp.w

Answers

To slide it up a 8 foot ramp since picking it up would be hard and a2 ft ramp is too short and the 4 ft ramp would make it harder to get it to the top for it will slide easier
The easiest way to move the box up is to slide it up a 4 foot ramp

A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?

Answers

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

[tex]\mu mg = m\frac{v^2}{r}[/tex]

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

[tex]\mu=0.40[/tex] is the coefficient of friction between the tires and the road

m is the mass of the car

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

[tex]v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s[/tex]

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

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Final answer:

On a rainy day, the maximum speed at which a curve with a radius of 150 m can be safely negotiated is 12.43 m/s. This calculation is based on the centripetal force provided by the friction between the tires and the road surface.

Explanation:

To determine the maximum speed at which the curve can be safely negotiated, we need to consider the centripetal force acting on the car. The centripetal force is provided by the friction between the tires and the road surface. The formula for centripetal force is Fc = mv²/r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.

On a rainy day, the coefficient of friction between the tires and the road is given as 0.40. The maximum frictional force is given by the equation Ff = μN, where μ is the coefficient of friction and N is the normal force.

Since the normal force is equal to the gravitational force acting on the car, we can calculate it using N = mg, where m is the mass of the car and g is the acceleration due to gravity.

To find the maximum speed, we need to equate the centripetal force and the frictional force: mv²/r = μN. We can rearrange this equation to solve for the maximum velocity v: v = sqrt(μrg).

Substituting the given values, v = sqrt(0.40 * 9.8 * 150) = 12.43 m/s.

Plants are producers. What is the role of plants in an ecosystem?

A.
Plants help animals make their own food.

B.
Plants are food for some consumers.

C.
Plants use oxygen to make their own food.

D.
Plants break down dead animals.


PLSSSSS HELP ME QUICK!

Answers

Answer:

(C) Plants use oxygen to make their own food

Answer:

B.

Explanation:

We can use process of elimination here.  A is not correct.  Animals do not make their own food.  C is not correct.  Plants use carbon dioxide, not oxygen to power photosynthesis.  D is not correct.  Decomposers break down dead animals.  Therefore, B is correct.  Plants are food for some consumers.  

Which of the following is a characteristic of sexually active teens?
Group of answer choices

single and step-parent families

poor school performance

same-sex parents

A and B are correct

A, B, and C are correct

Answers

Out of all given options, poor school performance  is a characteristic of sexually active teens.

Answer: Option B

Explanation:

Socio-demographic characteristics include age, gender and the highest level of education of the respondent. Most respondents had mothers with completed high school levels or diploma. Just over a quarter of respondents did not know their mother's highest education.  A low level of education and some family situations made the teen to be sexually active.

Teen sexual activity is "a fact in most African countries, as in most Western societies." There is a relationship between the sexual activity of young girls and their school resignation. Therefore, additional consequences of early sexual intercourse are women (and sometimes men) who drop out of school when they become pregnant.

Final answer:

Sexually active teens are more likely to come from single and step-parent families and have poor school performance.

Explanation:

The characteristic of sexually active teens is A and B are correct, which means they are more likely to come from single and step-parent families and have poor school performance.

While there may be studies suggesting correlations between certain factors, making broad statements about the behavior of sexually active teens based solely on family structure and school performance can be misleading.

Teenage sexual behavior is influenced by a variety of factors, including individual, family, cultural, and socioeconomic aspects. Family structure alone does not determine a teen's likelihood of engaging in sexual activity. Additionally, attributing poor school performance solely to family structure oversimplifies the complex interplay of factors influencing academic achievement.

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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.


a) Find the acceleration of gravity (g) at 2.60 x 10^6 m above the Earth.

b) Find the critical velocity to maintain a circular orbit at 2.60 x 10^6 m above the surface of Earth. (in both m/sec & km/hr)

c) Find the period of the orbit.

d) How many orbits will the satellite make in one day?

e) Find the escape velocity for a satellite to leave this orbit and escape the gravitational pull of the Earth.

f) Find the escape velocity for a rocket leaving the surface of the Earth.

Answers

a) The acceleration of gravity is [tex]4.96 m/s^2[/tex]

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

[tex]g=\frac{GM}{(R+h)^2}[/tex]

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)

[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)

[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)

Substituting,

[tex]g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2[/tex]

b)

The critical velocity for a satellite orbiting around a planet is given by

[tex]v=\sqrt{\frac{GM}{R+h}}[/tex]

where we have again:

[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)

[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)

[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)

Substituting,

[tex]v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s[/tex]

Converting into km/h,

[tex]v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h[/tex]

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

[tex]T=\frac{2\pi (R+h)}{v}[/tex]

where

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=2.60\cdot 10^6 m[/tex]

v = 6668 m/s

Substituting,

[tex]T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s[/tex]

d)

One day consists of:

[tex]t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s[/tex]

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

[tex]n=\frac{t}{T}=\frac{86400}{8452}=10.2[/tex]

e)

The escape velocity for an object in the gravitational field of a planet is given by

[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]

where here we have:

[tex]M=5.98\cdot 10^{24} kg[/tex]

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=2.60\cdot 10^6 m[/tex]

Substituting, we find

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s[/tex]

f)

We can apply again the formula to find the escape velocity for the rocket:

[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]

Where this time we have:

[tex]M=5.98\cdot 10^{24} kg[/tex]

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=0[/tex], because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s[/tex]

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If the box weighs 1,500 N, how much work dose the force of gravity do on the box

Answers

Answer:

The work done by the force of gravity, W = 0 J

Explanation:

Given data,

The weight of the box, w = 1500 N

Therefore the gravitational force acting on the box is , F = 1500 N

The displacement of the box S = 0

The work done is defined as the product of force acting on the body to the displacement it caused.

The formula for work done is,

                                           W = F x S

                                                = 1500 N x 0

                                                = 0

Hence. the work done by the force of gravity is, W = 0 J

Final answer:

The work done by the force of gravity on a box weighing 1500 N depends upon the distance the box is moved. If we assume a free fall situation, the work done would be calculated using the formula: W = 1500N * h, where h is the height from which the box falls.

Explanation:

The calculation of the work that the force of gravity does on an object comes down to two important factors: the force of gravity acting on the object, represented as weight, and the distance the object is moved. In this case, the box weighs 1500 N, which is the force of gravity acting on it. If we assume that the box is in free fall, i.e., it's falling to the ground under gravity, we can calculate the work done by the force of gravity.

To calculate this, we need to know the height from which the box is falling (let's denote that as h). The work done, according to the work-energy principle, is equal to the force times the distance covered and the cosine of the angle between force and distance vectors. Considering the angle between the gravity force and the displacement is 0 degrees in a free fall, the work done by gravity is W = 1500N * h * cos(0) = 1500N * h. Here, cos(0) is equal to 1.

As you can see, the work done by gravity relies heavily on the distance/height textbox falls. If the box doesn't move, then no work is done by gravity even though the force of gravity exists, because work is dependent on displacement.

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If the value of g=9.8;R=6400;G=6.67x10^-11then what is the value of M.
FORMULA =g=GM/R^2
Or gR=Gm/R

Answers

Answer:

The mass of the earth is, M = 5.98 x 10²⁴ kg

Explanation:

Given data,

The acceleration due to gravity, g = 9.8 m/s²

The radius of Earth, R = 6400 km

                                    =6.4 x 10⁶ m

The universal gravitational constant, G = 6.67 x 10 ⁻¹¹ Nm²kg⁻²

Using the equation,

                         GM = gR²

                            M = gR² / G

Substituting the values,

                            M = 9.8 x (6.4 x 10⁶)² / 6.67 x 10 ⁻¹¹

                                = 5.98 x 10²⁴ kg

Hence, the mass of the earth is, M = 5.98 x 10²⁴ kg

An electromagnet can be made by wrapping wire around which object?​

Answers

Answer:

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of the velocity? (1 point) b. What is the vertical component of the velocity? (1 point) c. How long does it take the water balloon to reach its highest point? (1 point) d. What is the maximum height of the water balloon? (1 point) e. What is the total amount of time that the water balloon is in the air? (1 point) f. How far does the water balloon land from where you launched it? (1 point)

Answers

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

[tex]v_x = u cos \theta[/tex]

where

u = 8.3 m/s is the initial velocity of the balloon

[tex]\theta=27^{\circ}[/tex] is the angle of projection

Substituting,

[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]

b)

The vertical component of the initial velocity of a projectile is given by

[tex]u_y = u sin \theta[/tex]

where

u is the initial velocity

[tex]\theta[/tex] is the angle of projection

Here we have

u = 8.3 m/s

[tex]\theta=27^{\circ}[/tex]

Substituting,

[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]

c)

The vertical component of the velocity of the balloon follows the suvat equation

[tex]v_y = u_y - gt[/tex]

where

[tex]v_y[/tex] is the vertical velocity at time t

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

[tex]v_y = 0[/tex]

So we get:

[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]

d)

The maximum height of the balloon can be calculated using the suvat equation:

[tex]s=u_y t - \frac{1}{2}gt^2[/tex]

where

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

[tex]t=\frac{2u_y}{g}[/tex]

where

[tex]u_y[/tex] is the initial vertical velocity

[tex]g[/tex] is the acceleration of gravity

Here we have

[tex]u_y = 3.8 m/s[/tex]

[tex]g=9.8 m/s^2[/tex]

Substituting,

[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

[tex]d=v_x t[/tex]

where

[tex]v_x[/tex] is the horizontal velocity

t is the time of flight

Here we have

[tex]v_x = 7.4 m/s[/tex]

[tex]t = 0.78 s[/tex]

Substituting,

[tex]d=(7.4)(0.78)=5.77 m[/tex]

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how does a pith ball electroscope work

Answers

Answer:

look at the explanation

Explanation:

This pith-ball electroscope is used to detect the presence of a static electricity charge. The two lightweight “pith” balls suspended from the strings are attracted to objects with a static electric charge. The pith balls can also be charged by touching them to an object with a static electric charge.

An ocean wave has a wavelength of 10 m and a frequency of 4.0 Hz. What is the velocity of the
wave? Show the appropriate equation from your book and show your work with units.

Answers

The velocity of the wave is 40 m/s

Explanation:

The relationship between the velocity of a wave and its frequency and wavelength is given by

[tex]v=f \lambda[/tex]

where

v is the velocity

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the ocean wave in this problem, we have:

[tex]\lambda = 10 m[/tex] (wavelength)

f = 4.0 Hz (frequency)

Therefore, its velocity is

[tex]v=(4.0)(10)=40 m/s[/tex]

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The velocity of an ocean wave with a wavelength of 10 m and a frequency of 4.0 Hz is 40 m/s.

The subject of this question is Physics, and more specifically, it deals with the topic of wave motion and wave velocity. The velocity of a wave can be calculated using the equation v = f × lambda, where v is the velocity of the wave, f is its frequency, and lambda is its wavelength.

Given that the wavelength lambda is 10 m and the frequency f is 4.0 Hz, we can substitute these values into the equation to find the wave velocity:

v = f × lambda = 4.0 Hz × 10 m = 40 m/s.

Therefore, the velocity of the ocean wave is 40 meters per second (m/s).

Which following is true off free falling object of different mass
A.they have the same acceleration but different g value
B.they have the same g value but different acceleration
C.they experience the same force of gravity
D.the rate of acceleration is the same for all objects

Answers

Answer:

Answer: C

explanation:

They could be same or different

ie:

|-5|,|-5| = 5,5

|-5|,|5| = 5,5

Answer:

C

Explanation:

9.8 m/s/s as well

The current theory of the structure of the
Earth, called plate tectonics, tells us that the
continents are in constant motion.
Assume that the North American continent
can be represented by a slab of rock 5100 kmon a side and 30 km deep and that the rock
has an average mass density of 2850 kg/m3
.
The continent is moving at the rate of about
3.8 cm/year.
What is the mass of the continent?
Answer in units of kg.

What is the kinetic energy of the continent?
Answer in units of J.

A jogger (of mass 80 kg) has the same kinetic
energy as that of the continent.
What would his speed be?
Answer in units of m/s.

Answers

1) Mass of the continent: [tex]2.2\cdot 10^{21} kg[/tex]

2) Kinetic energy: 1118 J

3) Speed of the jogger: 5.3 m/s

Explanation:

1)

First of all, we calculate the volume of the continent. It is a slab of side

[tex]L=5100 km = 5.1\cdot 10^6 m[/tex]

and thickness

[tex]t=30 km = 3.0\cdot 10^4 m[/tex]

So its volume is

[tex]V=tL^2=(3.0\cdot 10^4)(5.1\cdot 10^6)^2=7.8\cdot 10^{17} m^3[/tex]

The density of the slab is

[tex]\rho = 2850 kg/m^3[/tex]

Therefore, we can calculate the mass using the relationship

[tex]\rho = \frac{m}{V}[/tex]

where m is the mass. And solving for m,

[tex]m=\rho V=(2850)(7.8\cdot 10^{17})=2.2\cdot 10^{21} kg[/tex]

2)

The kinetic energy of the continent is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

[tex]m=2.2\cdot 10^{21} kg[/tex] is its mass

v = 3.8 cm/year is its speed

We have to convert the speed into m/s. Keeping in mind that

1 cm = 0.01 m

[tex]1 year = 365\cdot 24\cdot 60 \cdot 60 =3.15\cdot 10^7 s[/tex]

We find

[tex]v=3.18 \frac{cm}{y} \cdot \frac{0.01}{365\cdot 24 \cdot 60 \cdot 60}=1.0\cdot 10^{-9} m/s[/tex]

So now we can find the kinetic energy:

[tex]K=\frac{1}{2}(2.2\cdot 10^{21})(1.0\cdot 10^{-9})^2=1118 J[/tex]

3)

The kinetic energy of the jogger is given by

[tex]K=\frac{1}{2}m'v'^2[/tex]

where

m' = 80 kg is the mass of the jogger

v' is the speed of the jogger

Here we want the jogger to have the same kinetic energy of the continent, so

[tex]K=1118 J[/tex]

And by re-arranging the equation, we can find what speed the jogger must have:

[tex]v'=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1118)}{80}}=5.3 m/s[/tex]

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Which type of energy is made by a door slamming?
mechanical energy
light energy
electrical energy
sound energy

Answers

Answer:

SOUND ENERGY is the energy which gets released or formed.

Explanation:

Remember , Energy is ALWAYS conserved. It can neither be destroyed nor be created(made).

Mechanical energy : When the door slams , it's kinectic energy reduces and the friction between the door and the door frame cause some heat energy to be lost. This energy is got from the mechanical energy of the door.Sound Energy : The mechanical energy of the door partly converts into sound energy in order to produce sound.

Energy gets converted significantly to sound energy - from the following options.

Final answer:

The type of energy made by a door slamming is sound energy, which is a result of the door's mechanical energy being partially transformed into pressure waves in the air.

Explanation:

When a door slams, it produces sound energy as a result of the forceful closure that creates pressure waves in the air we perceive as sound. While the door itself experiences mechanical energy, which is necessary for the motion of slamming, the type of energy being asked about in this case is the one that is made as a consequence of that action, which is sound energy. Mechanical energy is the sum of kinetic and potential energy within a system and is responsible for physical movement, as when a person pushes a door causing it to slam. Upon impact, part of this mechanical energy is transformed into sound energy.

A 88.00 kg man is swinging on a rope, and the rope has a length of 12.0 meters. The mans speed at the bottom of his swing is 10.20m/s. If the swing can be treated like an object moving in a vertical circle, what is the maximum amount of tension in the rope during the swing?

Answers

Final answer:

The maximum tension in the rope during the swing is the sum of the gravitational force and the centripetal force, which can be calculated with the given mass, rope length, and speed, resulting in a tension of 1602.48 N.

Explanation:

To calculate the maximum tension in the rope during the swing, we need to consider the forces acting on the man at the lowest point of the swing. These forces are the gravitational force (weight) and the centripetal force required to maintain the circular motion. The tension in the rope is the sum of these two forces.

The gravitational force (Fg) on the man can be calculated using the equation Fg = m  imes g, where m is the man's mass (88.00 kg) and g is the acceleration due to gravity (9.81 m/s2). The centripetal force (Fc) required for circular motion can be calculated using the equation Fc = m imes v2 / r, where v is the man's speed (10.20 m/s) at the bottom of the swing and r is the radius of the swing, which is equal to the length of the rope (12.0 meters).

Therefore, the maximum tension (T) in the rope would be:

T = Fg + Fc

T = (88.00 kg  imes 9.81 m/s2) + (88.00 kg  imes (10.20 m/s)2 / 12.0 m)

Calculating each part:

Fg = 863.28 N (2 decimal places)

Fc = 739.20 N (2 decimal places)

Thus, the maximum tension in the rope would be:

T = 863.28 N + 739.20 N

T = 1602.48 N (rounded to two decimal places)

Tension, centripetal force, and gravitational force are crucial components in determining the tension in the rope during circular motion.

13. If both the directions of velocity and acceleration are negative, ...........
a velocity of the object increases
6 velocity of the object decreases
velocity of object is constant
the object stops motion .. Explain please​

Answers

Answer:

B)velocity of object decreases

Explanation:

Consider the positive x axis as positive direction

Assume a body moving in negative x-axis direction

It's acceleration also alone negative x-axis direction

So according to our consideration

velocity and acceleration values are negative

That is both are towards negative x direction

But as both velocity and acceleration are in same direction, MAGNITUDE of velocity increases

But as magnitude increases in negative direction, velocity value decreases

But speed value increases(As speed is scalar and velocity is a vector)

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise
the pitch to C sharp (271 Hz)? Use 343 m/s as the speed of sound.
a. 2.8 m/s
c. 11.4 m/s
b. 15.6 m/s
d. 8.2 m/s

Answers

Answer:

The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

[tex]\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}[/tex]

[tex]\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}[/tex]

[tex]f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)[/tex]

[tex]\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)[/tex]

[tex]\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

[tex]v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

Substitute the given values in the formula,

[tex]v_{s}=343+\frac{262}{271}(343-0)[/tex]

[tex]v_{s}=343+0.966(343)[/tex]

[tex]v_{s}=343-331.33[/tex]

[tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Therefore, The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

100 points, will mark Brainliest, please show work (As I'd like to learn how to solve this on my own in the future :^)

"With other strict requirements, an astronaut cannot be taller than 6 foot, 3 inches or have a mass greater than 95 kilograms. Determine the weight of the most massive astronaut on the moon where the gravity is one-sixth that of the Earth."

Answer choices:

120
140
160
180

Thanks in advance!

Answers

Answer:

180

Explanation:

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second. The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius.

Answer:

160kg

Explanation:

Given m= 95kg g= 1.6345m/s²

Unknown Fg= ?

Formula Fg=mg

Fg=(95)(1.6345)

Fg= 155.2775

Fg=160kg

The speed in case a is twice that of case b. The circles radius in case a is two times that of case b. How does the acceleration in case a compare to the acceleration in case b?

Answers

Final answer:

The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.

Explanation:

In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.

In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.

Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.

Learn more about Centripetal Acceleration here:

https://brainly.com/question/14465119

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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.

Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:

|a| = |v|²/r

From the problem, we have two cases:

[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]

Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]

For case A:

[tex]|a_A| = |v_A|^2 / |r_A|[/tex]

For case B:

[tex]|a_B| = |v_B|^2 / |r_B|[/tex]

Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:

[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]

Simplifying this, we get:

[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]

Therefore, the acceleration in case A is twice that of case B.

a soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill

Answers

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

A force F~ = Fx ˆı + Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 8 N, Fy = −1 N, sx = 6 m, and
sy = 2 m.
Find the work done by the force on the
particle.
Answer in units of J.

Answers

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

[tex]tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\[/tex]

Then we find the angle β of the displacement vector

[tex]tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\[/tex]

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

[tex]W=F*d*cos (theta)[/tex]

The absolute value of F will be:

[tex]F=\sqrt{8^{2}+1^{2}  } \\F= 8.06 N[/tex]

The absolute value of d will be:

[tex]d=\sqrt{(6 )^{2}+(2)^{2}  } \\d= 6.32m\\[/tex]

Now we have:

[tex]W=8.06*6.32*cos(25.56)\\W=46 J[/tex]

Spaceship 1 and spaceship 2 have equal masses of 300 kg. spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 6 m/s. they collide and stick together. what is their speed?

Answers

In a collision between two 300 kg spaceships, one at rest and the other moving at 6 m/s, the final velocity after they stick together is 2 m/s.

When two objects with masses of 300 kg collide and stick together, their momentum is conserved. Utilizing the principle of conservation of momentum, we can calculate their final velocity after the collision.

In this case, since both spaceships have equal masses and one is at rest while the other moves with a known speed, the final velocity after the collision would be 2 m/s.

3. A roller coaster car at the top of a hill, right before it plunges downward
has
a. potential energy
b. kinetic energy.
c. momentum
d. electrical energy.

Answers

The correct answer is a. potential energy. At the top of a hill, right before it plunges downward, a roller coaster car has potential energy due to its position relative to the ground. This potential energy is converted into kinetic energy as the car moves downward and gains speed.

a .101 kg particle undergoes simple harmonic motion along the horizontal x axis between the points x1 = - .219 m and x2 = .495 m. the period of oscillation is .513 s find the frequency, f, the equilibrium position, xep, the amplitude, a, the maximum speed, v max, the maximum magnitude of acceleration, a max, the force constant, k, and the total mechanicalenergy, e tot.​

Answers

1) Frequency: 1.95 Hz

2) Equilibrium position: 0.138 m

3) Amplitude: 0.357 m

4) Maximum speed: 4.36 m/s

5) Maximum acceleration: [tex]53.1 m/s^2[/tex]

6) Spring constant: 15.0 N/m

7) Total mechanical energy: 0.96 J

Explanation:

1)

The frequency of a simple harmonic motion is equal to the reciprocal of the period:

[tex]f=\frac{1}{T}[/tex]

where

f is the frequency

T is the period

For the particle in this problem, we have

T = 0.513 s (period)

So the frequency of motion is

[tex]f=\frac{1}{0.513}=1.95 Hz[/tex]

2)

In a simple harmonic motion, the object oscillates between two maximum positions [tex]+x_m[/tex] and [tex]-x_m[/tex] which are equidistance from the equilibrium position. So, the equilibrium position is the midpoint between these two positions.

For the particle in this problem, the two extreme positions are:

[tex]x_1 = -0.219 m[/tex]

[tex]x_2 = 0.495 m[/tex]

So the mid-point (the equilibrium position) is

[tex]x_{ep} = \frac{x_1 + x_2}{2}=\frac{-0.219+0.495}{2}=0.138 m[/tex]

3)

The amplitude of a simple harmonic motion is the maximum displacement of the object, measured from the equilibrium position.

This means that we can calculate the amplitude simply as the difference between one of the extreme positions and the equilibrium position.

Taking

[tex]x_2 = 0.495 m[/tex]

and

[tex]x_{ep} = 0.138 m[/tex]

We find the amplitude:

[tex]A=x_2 - x_{ep} = 0.495-0.138 =0.357 m[/tex]

4)

In a simple harmonic motion, the maximum speed is given by

[tex]v_{max}=\omega A[/tex]

where

[tex]\omega[/tex] is the angular frequency

A is the amplitude

The angular speed can be calculated from the frequency as follows:

[tex]\omega=2\pi f=2 \pi (1.95 Hz)=12.2 rad/s[/tex]

The amplitude is

A = 0.357 m

So, the maximum speed is

[tex]v_{max} = (12.2)(0.357)=4.36 m/s[/tex]

5)

The maximum acceleration in a simple harmonic motion is given by

[tex]a_{max}= \omega^2 A[/tex]

Where we already know that:

[tex]\omega=12.2 rad/s[/tex] is the angular frequency

A = 0.357 m is the amplitude of motion

Substituting, we find the maximum acceleration:

[tex]a_{max}=(12.2)^2(0.357)=53.1 m/s^2[/tex]

6)

The angular speed in a simple harmonic motion can be written as

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here we know that:

[tex]\omega=12.2 rad/s[/tex] is the angular speed

m = 0.101 kg is the mass of the particle

So we can solve the formula for k, the spring constant:

[tex]k=\omega^2 m =(12.2)^2(0.101)=15.0 N/m[/tex]

7)

Since the energy is conserved, the total mechanical mechanical energy is just equal to the maximum potential energy of the system, which occurs when the particle is at maximum displacement (x=A) and the speed is zero (so the kinetic energy is zero), therefore it is given by:

[tex]E=\frac{1}{2}kA^2[/tex]

where

k = 15.0 N/m is the spring constant

A = 0.357 m is the amplitude

And substituting,

[tex]E=\frac{1}{2}(15.0)(0.357)^2=0.96 J[/tex]

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Which one of the above forms of em radiation does our body detect as heat?

Answers

Answer: Infrared radiation

Explanation:

Infrared radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm), however this type of electromagnetic radiation is detected by the human body as heat.

 

These waves can be divided into:

- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.  

- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.  

- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.  This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the horizontal. The magnitude of the friction force acting on the block is:

Answers

The force of friction is 25 N

Explanation:

For this problem, we can apply Newton's second law of motion along the horizontal direction:

[tex]\sum F_x = ma_x[/tex]

where

[tex]\sum F_x[/tex] is the net force in the horizontal direction

m is the mass of the block

[tex]a_x[/tex] is the horizontal acceleration

Here  the block is moving at constant speed, so its acceleration is zero, therefore:

[tex]a=0 \rightarrow \sum F_x = 0[/tex] (1)

The net force in the horizontal direction can be written as:

[tex]\sum F_x = Fcos \theta -F_f[/tex] (2)

where

[tex]Fcos\theta[/tex] is the horizontal component of the pulling force, with F = 50 N being the magnitude and [tex]\theta=60^{\circ}[/tex] being the direction, acting forward[tex]F_f[/tex] is the force of friction, acting backward

Combining (1) and (2), we find the magnitude of the force of friction:

[tex]Fcos \theta -F_f = 0\\F_f = F cos \theta =(50)(cos 60^{\circ})=25 N[/tex]

Learn more about friction:

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2. (06.09A) Which expression is equivalent to one over five m − 20? (5 points) one over five (m − 4) one over five (m − 100) 5(m − 4) 5(m − 100)

Answers

Answer:

A

Explanation:

Answer:

a

Explanation:

a

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