In ultimate the disc may be passed in any direction.
True
False

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

F =  3.32 x 10⁻⁶ N

Answer:

Fall

Explanation:

I would say food but like they take weight too

We are given:

Homer the Astronaut:

Mass of Homer the astronaut(m1) = 200 kg

initial velocity of Homer the astronaut(u1) = 5 m/s

Larry the Pal:

Mass of Larry the Pal (m2) = 150 kg

initial velocity of Larry the Pal (u2) = 0 m/s

Since they will move together after the collision, they will have the same velocity:

v1 = v2 = V

Solving for the Final velocity:

from the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

since v1 = v2 = V:

m1u1 + m2u2 = V(m1 + m2)

replacing the variables with the given values

200 * 5 + 150 * 0 = V(200 + 150)

1000 = 350V

V  = 1000 / 350

V = 2.86 m/s

Answer:

y becomes doubled.

Explanation:

If;

       y  = [tex]\frac{k}{x}[/tex]  

what is the state of y when x is halved;

  the given expression is an inverse relationship. When y increases, x is supposed to decrease and vice versa.

 if x is halved; x  = [tex]\frac{x}{2}[/tex]

           [tex]\frac{k}{\frac{x}{2} }[/tex]   = [tex]\frac{2k}{x}[/tex]  

Now compare :

      [tex]\frac{k}{x}[/tex]  :  [tex]\frac{2k}{x}[/tex]

we see that y becomes doubled

Answer:

The average speed is 22.2 km/h

Explanation:

Average Speed

Given an object travels a total distance d and took a total time t, then the average speed is:

[tex]\displaystyle \bar v=\frac{d}{t}[/tex]

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]

Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

The average speed is 22.2 km/h

Answer:

+1

-1

Explanation:

The oxidation number of Na in NaCl is +1 and that of Cl is -1.

When NaCl is oxidized we have;

        NaCl →  Na⁺  + Cl⁻

In this process, Na will lose an electron to become Na⁺ and thus isoelectronic with Ne, Cl will gain the electrons to become Cl⁻ and thus isoelectronic with Argon.

Answer:

How long do you jump (sec) ?

Explanation:

Answer:

b

Explanation:

Answer:

they are cooler than the rest if the sun

Answer:

Poetic

Explanation:

I took the test

Answer:

Poetic

Explanation:

I got it right on the test

Answer:

a = 0.62 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.

[tex]v_{f} =v_{o} -a*t[/tex]

[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]

where:

Vf = final velocity = 2,77 [m/s]

Vo = initial velocity = 15.27 [m/s]

t = time = 20 [s]

a = acceleration [m/s²]

Now replacing:

[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]

Answer:

It would be 10.00

Explanation:

Hope this helps its different for everyone what was it for u it was D for me

Answer:

Explanation:

all biotic and abiotic factors around a living organism is its environment

Answer: An environment is the surroundings or conditions in which a person, animal, or plant lives or operates. the natural world, as a whole or in a particular geographical area, especially as affected by human activity. The best synonym for environment would be nature.

Explanation: Environment is a place where different things are such as a swampy or hot environment. They constantly interact with it and adapt themselves to conditions in their environment. In the environment there are different interactions between animals, plants, soil, water, and other living and non-living things. Environment plays an important role in healthy living and the existence of life on planet earth. Earth is a home for different living species and we all are dependent on the environment for food, air, water, and other needs. Therefore, it is important for every individual to save and protect our environment. To divide environments' sorts we can mention 3 kinds of environments Natural, industrial, and social environment. Natural environment: Include water, light, land, air and all organisms that live in nature.

Answer:

dont die bro

Explanation:

Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

Given:

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

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

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

Learn more about work done here:

https://brainly.com/question/24230840

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

read more here https://brainly.com/question/18103369

The force required to hold it completely submerged under water is 0.25 N

The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³

The density of water [tex]\rho_w[/tex] = 1000 kg/m³

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]

F = 0.25 N

Find out more at: https://brainly.com/question/19054414

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

Learn more: https://brainly.com/question/8155254

Answer:

The electrostatic force will remain the same

Explanation:

From the question we are told that  

     The charge on the each conducting sphere is [tex]q = 2 \ C[/tex]

      The radius each sphere is [tex]r_1 = r_2 = r = 0. 1\ m[/tex]

Generally electrostatic force between the sphere is mathematically represented as

             [tex]F = \frac{k * q_1 q_2 }{d^2}[/tex]

Here  k is the coulomb constant ,

         d  is the distance between the two sphere which is measured from one center of the sphere to the other center of the sphere  

Now from the question we are told that the radius of the spheres is doubled (i.e  from 0.10 m  to 0.2 m ) , this will not affect the distance between the sphere because their center are still in the same position

and given there is no change in the distance between the spheres , the electrostatic force will remain the same

Answer:

Th average force impact is [tex]F = 168.298 \ N[/tex]

Explanation:

From the question we are told that  

   The mass of the golf ball is  [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]

    The angle made is [tex]\theta = 33.6 ^o[/tex]

    The range of the golf ball  is [tex]R = 190 \ m[/tex]

     The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]

Generally the range of the golf ball is mathematically represented as

       [tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]

Here v  is the velocity with which the golf club propelled it with, making  v the subject

       [tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]

=>     [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]

=>     [tex]v = 44.94 \ m/s[/tex]

Generally the change in momentum of the golf ball is mathematically represented as

      [tex]\Delta p = m * (v - u )[/tex]

here u  is the initial  velocity of the ball before being stroked and the value is 0 m/s

       [tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]

=>    [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]

Generally the  average force of impact is mathematically represented as      

         [tex]F = \frac{\Delta p }{\Delta t}[/tex]

=>        [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]

=>        [tex]F = 168.298 \ N[/tex]

Answer:

yeah

Explanation:

yeah yeah yeah yeah

Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s

The formula for finding variables in a  Venturi tube is shown below:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity

P₁ - P₂ = 1.2 KPa = 1200 Pa

ρ = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

We then substitute the variables into this equation.

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²  ------ equation (1)

The  continuity equation A₁V₁ = A₂V₂ is then used

where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

We then substitute the value into equation 1

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

This can be calculated using the formula

Flow Rate = Q = A₂V₂

                  = (2.746 x 10⁻⁴ m²)(1.86 m/s)

                 = 5.1 x 10⁻⁴ m³/s

Answer:

the focal length is 4cm

Explanation:

use the formula 1+1/m = f

it is obtained from the mirror formula

m is the magnification and is the ratio of the image height to the object height

however they should have the same unit . that means you would have to do a little conversion from cm to mm or vice versa .

Answer:

(a) The time taken for the plane to reach 50 km/hr is 811.97 s

(b) (i) the acceleration of the plane is 0.012 m/s²

    (ii) The time taken for the plane to reach 50 km/hr is 1172.85 s

Explanation:

Given;

mass of the freight train, m = 3.8 x 10⁷ kg

force applied to pull the train, F = 6.5 x 10⁵ N

a. How long does it take the train to go from rest to 50 km/hr?

Given;

initial velocity, u = 0

final velocity, v = 50 km/hr = 13.889 m/s

Apply Newton's second law of motion to determine the time traveled by the train at the given speed;

[tex]F = ma = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{6.5*10^5}\\\\t = 811.97 \ s[/tex]

(b)

(i) the acceleration of the plane

Given;

air resistance, = 2.0 x 10⁵ N

Apply Newton's second law of motion;

∑F = ma

6.5 x 10⁵ N - 2.0 x 10⁵ N = ma

4.5 x 10⁵ N = ma

[tex]a = \frac{4.5 * 10^5 }{m}\\\\a = \frac{4.5 * 10^5 }{3.8*10^7}\\\\a = 0.012 \ m/s^2[/tex]

(ii) time taken for the plane to reach 50 km/hr is given by;

[tex]F = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{4.5*10^5}\\\\t = 1172.85 \ s[/tex]

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

big bang theory

Explanation:

Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.

It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all  it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.

Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.

Learn more here: https://brainly.com/question/24439982

Answer:

The total thermal energy required  is 8.93 Kcal

Explanation:

Given;

mass of the ice cube, m = 43 g

specific heat capacity of water, Cp = 4.18 J/gc

specific latent heat of fusion of ice, Cf = 334 J/g

First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C

Q₁ = mCp[0 - (-16.5)]

Q₁ = 43 x 4.18(16.5)

Q₁ = 2965.71 J

Second step, determine the latent heat of fusion of ice at 0°C

Q₂ = mCf

Q₂ = 43 x 334

Q₂ = 14362 J

Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.

The final temperature of the water = 11.5 °C + 100° C = 111.5 °C

Q₃ = mCp Δθ

Q₃ = 43 x 4.18 (111.5 - 0)

Q₃ = 20041 J

Total thermal energy required = Q₁ + Q₂ + Q₃

Total thermal energy required =  2965.71 J + 14362 J +  20041 J

Total thermal energy required = 37,368.71 J

Total thermal energy required = 8.93 Kcal