A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time interval, the car traveled ________ meters.

Answer:

Newtons 2nd Law of motion affects football as well. This law states the greater the mass of an object, the less the acceleration. This means that the more the objects weighs, the more difficult it is to move the object.

Explanation:

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Answer:

c

Explanation:

Answer:

this is about momentum p=mv

A, the elephant has more mass

Answer:

here

Explanation:

Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

Examples of conductors include metals, aqueous solutions of salts

Answer:

Explanation:

The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.

Answer:

C. zero for both

Explanation:

In case of solid metal sphere , when it is given any charge , all the charges are transferred on the surface and within surface no charge exists . In case of hollow metal sphere , all charges reside on surface . In this way , in both solid and hollow sphere , all charge reside on the surface and no charge resides inside it . Hence due to absence of any charge inside  , there is no electric field inside the sphere in both the cases .  

Hence in both the case electric field is zero .

option C is correct .

Explanation:

C. meter per second squared

B. meter per second

A. meter

Answer:

b. meters per second

c.meters per second squared

c.meters

Explanation:

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Answer:

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Answer:

ye ek rod h or electric ⚡ field h P point

Answer:

- time t taken for car to travel is 64.57 s

- distance travelled between A and B is 1.4887 km

Explanation:

Given the data in the question;

[tex]U_{BC}[/tex] =  83 km/h = ( 83×1000 / 60×60) =  23.0555 m/s

[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) =  11.3888 m/s

now, we calculate the acceleration;

a =  (  [tex]U_{BC}[/tex] -  [tex]U_{CD}[/tex] ) / t

we substitute

a =  ( 23.0555 -  11.3888 ) / 4.4

a = 11.6667 / 4.4

a = 2.6515 m/s²

Now equation for displacement from BC

[tex]S_{BC}[/tex] =  [tex]U_{BC}[/tex]t + 1/2.at²

we substitute

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

[tex]S_{BC}[/tex] = 75.7792 m

Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m

so

[tex]S_{AB}[/tex] +  [tex]S_{BC}[/tex] +  [tex]S_{CD}[/tex]  =  2300 m

we substitute substitute

[tex]S_{AB}[/tex] +  75.7792 m +  [tex]S_{CD}[/tex]  =  2300 m

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex] = 2300 - 75.7792

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex]  = 2224.2208 m

so we substitute 23.0555t for [tex]S_{AB}[/tex]  and 11.3888t for  [tex]S_{CD}[/tex]  

23.0555t + 11.3888t  = 2224.2208

34.4443t = 2224.2208

t = 2224.2208 / 34.4443

t = 64.57 s

Therefore, time t taken for car to travel is 64.57 s

Distance Between A to B

[tex]S_{AB}[/tex]  = t ×  [tex]U_{AB}[/tex]

we substitute

[tex]S_{AB}[/tex]  = 64.57 s × 23.0555

[tex]S_{AB}[/tex]  = 1488.69 m

[tex]S_{AB}[/tex]  = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km

Answer:

Acceleration due to gravity, g = 2.68m/s²

Explanation:

Given the following data;

Period = 5.14s

Length = 0.25m

To find acceleration due to gravity, g;

[tex] Period, T = 2 \pi \sqrt {lg} [/tex]

Substituting into the equation, we have;

[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]

[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]

[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]

[tex] 0.8180 = \sqrt {0.25g} [/tex]

Taking the square of both sides

[tex] 0.8180^{2} = 0.25g [/tex]

[tex] 0.6691 = 0.25*g[/tex]

[tex] g = \frac {0.6691}{0.25} [/tex]

Acceleration due to gravity, g = 2.68m/s²

Answer:

a=1m/s^2

Explanation:

1÷1÷1=1m/s^2

Answer:

 t = 14.53 s

Explanation:

The speed of a wave is constant and is given by

         v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]

in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations

          v = x / t

           t = x / v

in this case the sound pulse leaves the ship and must return so the distance is

          x = 2d

where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s

let's calculate

           t = 2 10900/1500

           t = 14.53 s

Answer:

2.18 miles per hour

Explanation:

Given:

A hiker walks 7.7 miles to the east in 3.6 hours.

A hiker walks 2.1 miles to the west in 2.4 hours

To find: magnitude of average velocity during the trip

Solution:

Total distance = 7.7 + 2.1 = 9.8 miles

Total Time = 2.1 + 2.4 = 4.5 hours

Average velocity = Total distance ÷ Total Time =[tex]\frac{9.8}{4.5}=2.18[/tex] miles per hour

Answer:

A) 1268 MeV

B)  299MeV/c

C) 1268 MeV

Explanation:

Given :

π meson rest energy = 139.6 MeV

Speed = 0.921c

proton at rest energy = 938.3 MeV

a) Find the total relativistic energy of resulting composite particle

E = E(meson) + E(proton)

  = [tex]\frac{(mc^2)_{meson} }{\sqrt{1-\frac{v^2}{c^2} } } + (mc^2)_{proton}[/tex]

  = [tex]\frac{139.6MeV}{\sqrt{1-\frac{(0.906c)^2}{c^2} } } + 938.3[/tex]

E = 1268 MeV

B) determine the total linear momentum of the composite particle

= 299MeV/c

attached below is the detailed solution

C) Determine the rest energy of the composite particle

E = 1268 MeV

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

Answer:

Explanation:

the center of mass formula

Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)

Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.

y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1

Ycm ( given ) = - .5

Putting the values of masses and positions

- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )

- .5 = 128  + 14  + m₃ x - 2.1 / ( 100+ m₃ )

- 50 - .5 m₃ = 142 - 2.1 m₃

1.6 m₃ = 192

m₃ = 120 kg .

B )

Total downward force is weight of  total mass  = 80 + 20 + 120

= 220 kg

weight = 220  x 9.8 = 2156 N .

component of weight perpendicular to rope

= 2156 cos 15 = 2082.53 N

This force will be equally distributed over each tree , so force on each tree =  2082.53 / 2 = 1041.26 N .

Answer:

Thats probably 11

Explanation:

Answer:

yes

Explanation:

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Explanation:

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

We have that the speed  must be at the speed below if they want to just hit their target

From the Question we are told that

Distance [tex]d=50m[/tex]

Height [tex]h=70m[/tex]

Constant Velocity [tex]v= 7.0 m/s[/tex]

Generally the equation for the time  is mathematically given as

[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]

Therefore

The velocity required to make horizontal movement is

[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]

Given that

Velocity on the Vertical axis is

[tex]v_y=7m/s[/tex]

Velocity on the  horizontal axis is

[tex]v_x=5m/s[/tex]

Therefore resultant speed

[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]

[tex]v_r=8.6023m/s[/tex]

In conclusion

[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target

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Answer:

θ = 225 rad

Explanation:

given data

angle = 25 rad

to find out

angular velocity after 3t?

solution

let angular acceleration α in t

θ = ω × t + 0.5 × α × t²        ........................1

here ω  = 0 (initial velocity )

so put this value here

25 = 0 + 0.5 × α × t²             ..........................2

α = 25 ÷ (0.5 t²)

α = 50 ÷ t²                            .........................3

now here we take in 3t

θ = ω × 3t + 0.5 × α × (3t)²

for ω  = 0

θ = 0 + 0.5 × α × 9t²

now put value in eq 2

so

θ = (0.5) ×  (50 ÷ t²) ×  (3t)²

θ = 25 × 9

θ = 225 rad

Answer:

ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

q = +9 10⁵ C

Explanation:

Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:

* ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

This creates that the planet has a net electrical load

We can roughly calculate the charge of the planet

             E = k q / r²

             q = E r² / k

let's calculate

             q = 200 (6.37 10⁶)²/9 10⁹

              q = +9 10⁵ C

Answer:

it has a higher amplitude than that of the original waves

Explanation:

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The two waves combined with constructive interference have a higher amplitude than the original wave.

High energy waves are characterized by high amplitude. Low energy waves are characterized by their small amplitude. As explained in Lesson 2, wave amplitude is the maximum amount of particle movement on a medium from a resting position.

Superposition is a combination of two waves in the same place. Constructive interference occurs when two identical waves interfere in phase. Destructive interference occurs when two identical waves are exactly out of phase and overlap.

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Answer:

Explanation:

Speed of sound in air at room temperature = 346 m /s

Let speed of sound in metal given = v m /s

Time take by sound wave travelling through air = 4 / 346 = .01156 s

Time taken by sound wave travelling through metal = 4 / v s

Given ,

.01156 - 4 / v = 11 x 10⁻³ s = .011 s ( time taken by sound travelling in air is more )

4 /v = .01156 - .011 = .00056

v = 4 / .00056 = 7142.85 m /s

Answer:

2.1m/s towards your grandmother's house

Explanation:

Given parameters:

Time taken  = 204s

Distance  = 430m

Unknown:

Velocity  = ?

Solution:

The velocity is determined by:

    Velocity  = [tex]\frac{displacement}{time}[/tex]  

   Velocity  = [tex]\frac{430}{204}[/tex]   = 2.1m/s towards your grandmother's house

Answer:

Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.

Answer:

d = 106.41 m

Explanation:

Given that,

Initial speed of the car, u = 60 mph = 26.9 m/s

The deceleration in the car, a = -3.4 m/s²

The average reaction time, t = 0.218 m/s

It finally stops, final velocity, v = 0

We need to find the distance covered by the car as it come to a stop.

Using third equation of motion to find.

[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]

So, the car will cover 106.41 m as it comes to a stop.

so we will use Newton's gravitational law :

gravitational acceleration = G*m/r^2  

G is the gravitational constant = G = 6.673×10^-11 N m^2 kg^-2  

after substitution :  

6.673×10^-11 * m / (1.15 x 10^6)^2 = 0.61

= 5.04575*10^-23 * m = 0.61

dividing over 5.04575*10^-23 :

m = 1.20894*10^22 kg

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Answer:

4

2

1

3

Explanation:

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