A force (5ỉ - ;)N moves an object from the point P (1,3) m to the point Q (3, 8) m.
The work (in )) done by the force is
Select one:
a. -10
O b.5
o c.-13
d. 8​

Answer:

The correct solution is "14.6875 kg".

Explanation:

Given values:

Force,

F = 47.0 N

Acceleration,

a = 3.20 m/s²

Now,

⇒ [tex]Force=Mass\times Acceleration[/tex]

or,

⇒       [tex]F=ma[/tex]

⇒    [tex]47.0=m\times 3.20[/tex]

⇒       [tex]m=\frac{47.0}{3.20}[/tex]

⇒           [tex]=14.6875 \ kg[/tex]

Answer: The correct statements are:

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

Explanation:

F=w/s

F=40,000j/330m

F=121.212N

The lens strength or lens power of contact lenses that m should be prescribed to correct the nearsightedness is 59.9 diopters.

The focal length of a lens is the distance from a lens to the focal point of that lens.

It is a measure of how strongly a lens converses or diverges light.

The power of a lens is the inverse of the focal length when given in metres.

Power of a lens = 1/focal length (m)

The focal length = 16.7 mm = 0.0167 M

Power of lens required = 1/0.0167 = 59.9 diopters

Therefore, the lens strength or lens power of contact lenses that m should be prescribed to correct the nearsightedness is 59.9 diopters.

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Answer:

22.4 cm

Explanation:

The total mechanical energy of the particle Etot = K + U where K = kinetic energy = 1/2mv² where m = mass of particle = 100 g =0.1 kg and v = speed of particle = 2 m/s and U = potential energy = 1/2kx² where k = spring constant and x = displacement

Etot = K + U

U = Etot - K

U = Etot - 1/2mv²

Since Etot = 0.45 J, substituting the values of the other variables into the equation, we have

U = Etot - 1/2mv²

U = 0.45 J - 1/2 × 0.1 kg × (2 m/s)²

U = 0.45 J - 1/2 × 0.1 kg × 4 m²/s²

U = 0.45 J - 0.1 kg × 2 m²/s²

U = 0.45 J - 0.2 J

U = 0.25 J

Now U = 1/2kx² and the angular frequency ω = √(k/m) ⇒ ω²m = k

So, U = 1/2ω²mx²

x² = 2U/ω²m

x = (√2U/m)/ω

since ω = 10 rad/s, substituting the values of the variables into the equation, we have

x = (√2U/m)/ω

x = (√2 × 0.25 J/0.1 kg)/10 rad/s

x = (√0.5 J/0.1 kg)/10 rad/s

x = (√5 J/kg)/10 rad/s

x = 2.24/10 m

x = 0.224 m

x = 22.4 cm

As we know larger the area of contact lesser the pressure. So, in order to reduce the pressure heavy vehicles have broad tyres to increase the area of contact with the ground. Heavy vehicles have broad tyres because broad tyres have large area of contact and less pressure on the ground.

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Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

[tex]-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm[/tex]

The image is in the glass.  

Answer:

100 cm

Explanation:

It is given that there are two loudspeakers that produces [tex]$\text{sound waves }$[/tex] along x-axis.

The maximum intensity of the sound is [tex]$\text{when the speakers are}$[/tex] at a distance of = 15 cm apart.

The sound intensity becomes zero when the separation between the speakers are increased and becomes 65 cm.

Therefore, the sound waves are in the phase, [tex]$\Delta x_1=15 \ cm$[/tex]

The sound waves are out of phase when [tex]$\Delta x_2=65 \ cm$[/tex]

Therefore,

[tex]$\Delta x_2 - \Delta x_1 = \frac{\lambda}{2}$[/tex]

[tex]$\lambda= 2(\Delta x_2 - \Delta x_1)$[/tex]

  = 2 (65 - 15)

  = 2 x 50

  = 100 cm

Hence the wavelength of the sound is 100 cm.

Answer:

Equations of motion relate the displacement of an object with its velocity, acceleration and time. s=vt where s is the displacement, v the (constant) speed and t the time over which the motion occurred. ...

Displacement with negative acceleration: s, equals, v, t, minus, one half, a, t, square...

Displacement with positive acceleration: s, equals, u, t, plus, one half, a, t, squared,s...

Velocity squared: v, squared, equals, u, squared, plus, 2, a, s,v2=u2+2as

Velocity: v, equals, u, plus, a, t,v=u+at

Answer:

[tex](x \times 1) = (400 \times 2.5) \\ x = 1000 \: newtons[/tex]

[tex]y = 0[/tex]

Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Explanation:

Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Answer: [tex]4.65\ m/s[/tex]

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

[tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]

So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]

Answer:

The spring constant is 173.12 N/m.

Explanation:

mass, m = 32 kg

Period, T = 2.7 s

let the spring constant is K.

Use the formula of period,

[tex]T = 2\pi\times \sqrt \frac{m}{K}\\\\2.7 =2\times 3.14\sqrt\frac{32}{K}\\\\K = 173.12 N/m[/tex]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency​ reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, f = 39 kHz

velocity of the bat, [tex]v_b[/tex] = 8.32 m/s

speed of sound in air, v = 340 m/s

Apply the doppler-effect principle to solve this problem.

The apparent frequency of sound striking the wall is calculated as;

[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]

The frequency reflected by the stationary wall to the bat is calculated as;

[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]

[tex]f_s\approx 41 \ kHz[/tex]

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Kaise h0 aap...

Hello yarr kha h0 sab k sab

Avvu

Mdhu sis

Naina sis

jonain

.....etc

they have rubber and meatal so its going to get hot of course cause it a engine pushing and the gasoline is just all coming together but damage wise I still it the protection on that part to keep it from damage and life risk

I hope this may help

wave velocity = frequency × wavelength

300 = frequency × 0.75

Frequency = 300÷0.75 = 400 HZ

Answer:

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}[/tex]

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles[/tex]

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Answer:

Two beneficial aspects of the last 10,000 years of mankind have been the development of writing and the development of civilized communities.

Thus, writing has allowed the recording of the different historical events that have taken place during the development of humanity, setting precedents for subsequent generations, allowing documenting the progress that humanity has been achieving. On the other hand, the development of civilized communities around urban systems has allowed for greater economic and social development on the part of human groups, which through cooperation have been acquiring an increasingly higher quality of life.

On the other hand, two negative aspects of these years have been the world overpopulation and the pollution of the planet.

Thus, the amount of the world's population has been getting closer and closer to the limit of our planet's carrying capacity, which could lead humanity to a point where resources are definitely scarce, causing famines and various catastrophes. On the other hand, the pollution arising from the development of industrial societies has notably undermined the health of individuals and has damaged their standard of living, while causing irreparable damage to the natural environment.

Answer:

the answer is the explaination

Explanation:

Present beach erosion prevention methods include sand dunes, vegetation, seawalls, sandbags, and sand fences. Based on the research conducted, it is evident that new ways to prevent erosion must be obtained. Each way that is currently used has extensive negative effects on beaches and their natural tendencies.

Answer:

They will construct strong sea walls with the intent to protect the coastline and infrastructure that could be affected by erosion.

Explanation:

Answer:

30 newtons

explanations

data given

mass=6kg

acceleration=5

f=m×a

6×5=30

Answer:

transparent

Explanation:

Answer:

w= p∆v 50000 ( 0.55-0.40) and calculate and you get it

The work done on the system by its environment is 7,500J.

To find the work done, we need to know about the work done on an ideal gas during isobaric process.

When the pressure is remained constant throughout a process, then the process is called isobaric process.

Here, Final volume= 0.55 m³

         Initial volume= 0.40 m³

         Pressure= 50,000 Pa

Work done = 50,000 × (0.55 - 0.40)

                  = 7,500 J

Thus, we can conclude that 7,500 J of  work is done on the system by its environment.

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Answer:

to be honest i dont know

Explanation:

^^

Answer:

false

Explanation:

You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest

Answer:

the correct answer is B

Explanation:

The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values ​​from zero to infinity.

This quantum number is placed as a coefficient in the quantum distribution.

In this case for phosphorus, the number is n = 3

the correct answer is B

Answer:

total time taken is 2.1389 seconds

Explanation:  

Given the data in the question and as illustrated in the diagram below;

They have measured the acceleration of the ball on the incline to be 2.5 m/s²

a = -2.5 m/s²    { negative because its acting downward }

from the diagram, Motion from B to C will be;

v² = u² + 2as

given that distance s = 1.1 m

we substitute

v² - u² = 2as

(0)² - u² =  2 × (-2.5 m/s ) × 1.1 m

-u² = -5.5

u = √5.5

so Initial speed; u = 2.345 m/s

to get the time, we use the equation of motion;

v = u + at

we substitute

0 = 2.345 m/s + ( -2.5 m/s² × t₁ )

2.5t₁ = 2.345

t₁ = 2.345 / 2.5

t₁ = 0.938 s

Now, from C to B

we determine t₂

s = vt₂ + [tex]\frac{1}{2}[/tex]at₂²

1.1 m = 0×t₂ + [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²

1.1 m = [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²

1.1 m = 1.25 m/s² × t₂²

t₂² = 1.1 m / 1.25 m/s²

t₂² = 0.88 s²

t₂ = √0.88 s²

t₂ = 0.938 s

Next, from B to A; t₃ will be;

s = ut₃ + [tex]\frac{1}{2}[/tex]at₃²

we substitute

0.7 m = (2.345 m/s)t₃ + [tex]\frac{1}{2}[/tex] × 2.5 m/s × t₃²

0.7 m = (2.345 m/s)t₃ + ( 1.25 m/s )t₃²

1.25t₃² + 2.345t₃ - 0.7 = 0

using; x = [ -b ± √( b² - 4ac ) ] / 2a

we substitute

x = = [ -2.345 ± √( (2.345)² - (4 × 1.25 × -0.7) ) ] / 2(1.25)

x = [ -2.345 ± √( 5.499 + 3.5 ) ] / 2.5

x =  [ -2.345 ± √( 8.999 ) ] / 2.5

x =  [ -2.345 ± 2.9998 ] / 2.5

x =  ( [ -2.345 - 2.9998 ] / 2.5 ) or ( [ -2.345 + 2.9998 ] / 2.5 )

x = -2.1379 or 0.2629

hence t₃ = 0.2629 s

∴ Total time = t₁ + t₂ + t₃

Total time = 0.938 s + 0.938 s + 0.2629 s

Total time = 2.1389 seconds

Therefore, total time taken is 2.1389 seconds