A helicopter is hovering at a constant height of 35 m. The upward lift force on the helicopter is 85500. What is the weight of the helicopter

Answer:

c

Explanation:

Answer:

a=1m/s^2

Explanation:

1÷1÷1=1m/s^2

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:

ginawa ko na rin KC toh

Answer:  a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.

Explanation:

An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth

          hope this helps :)

Answer:

650

Explanation:

use the equation

weight = gm

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

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:

here

Explanation:

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

Examples of conductors include metals, aqueous solutions of salts

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:

16

Explanation:

[tex] \frac{1}{2} \times 2 \times {4}^{2} = 16[/tex]

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: light from the object

Explanation:

When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light

Answer:

     l = 0.548 m

Explanation:

For this exercise we compensate by finding the speed of the car

         p = m v

         v = p / m

         v = 0.58 / 0.2

         v = 2.9 m / s

this is how fast you get to the ramp, let's use conservation of energy

starting point. Lowest point

         Em₀ = K = ½ m v²

final point. Point where it stops on the ramp

         [tex]Em_{f}[/tex] = U = m g h

  mechanical energy is conserved

          Em₀ = Em_{f}

          ½ m v² = m g h

           h = [tex]\frac{m v^2}{2 g}[/tex]

let's calculate

          h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]

          h = 0.0858 m

to find the distance that e travels on the ramp let's use trigonometry, we look for the angle

          tan θ = y / x

          tan θ = 12/75 = 0.16

          θ = tan⁻¹ 0.16

          θ = 9º

therefore

           sin 9 = h / l

           l = h / sin 9

           l = 0.0858 / sin 9

           l = 0.548 m

Answer:

this is about momentum p=mv

A, the elephant has more mass

Answer:

The second option - the papers are acted upon by an unbalanced force from the fan.

Explanation:

Answer:

it has a higher amplitude than that of the original waves

Explanation:

trust me its right

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.

Learn more about higher amplitude waves at

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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:

yes

Explanation:

Answer:yeeeeeeeeeeeeeeeeeeeeeees

Explanation:

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:

a. 2500 cm³.

b. 2.5 litres.

Explanation:

Given the following data:

Density = 0.8g/cm³

Mass = 2000g

To find the volume of the petrol;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

Making volume the subject of formula, we have;

[tex]Volume = \frac{mass}{density}[/tex]

Substituting into the equation, we have:

[tex]Volume = \frac{2000}{0.8}[/tex]

Volume = 2500 cm³

a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.

b. The volume of the petrol in the can in litres;

1000 cm³ = 1 litre

2500 cm³ = x litres

Cross-multiplying, we have;

1000x = 2500

x = 2500/1000

x = 2.5 litres.

Therefore, the volume of the petrol in the can in litres is 2.5.

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:

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:

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Answer:

ye ek rod h or electric ⚡ field h P point

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 .

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

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:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

Answer:

84.1 m

Explanation:

Given :

The distance from the ship to submarine :

20 + y

Using Pythagoras :

Tan θ = opposite / Adjacent

Tan θ = 20 / 12

12 tan θ = 20

θ = tan^-1(20/12)

20

θ = 59.036°

The angle phi;

n1sinθ1 = n2sin θ

Sin 59.036 = 1.33 * sin phi

Sin phi = sinsin(59.04) ÷1.33

0.8574907 = 1.33 * sin phi

Sin phi = 0.8574907 / 1.33

Sin phi = 0.6447298

phi = sin(0.6447298

Phi = 40.15°

From Pythagoras :

y = 76tan40.15°

y = 76 * 0.8435707

y = 64.11

20 + y

20 + 64.11 = 84.11

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