whats 100+50+89+2000☹️

Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1  = 100kg

Mass of object 2 = 300kg

Distance  = 2m

Unknown:

Force of gravitational attraction between the objects  = ?

Solution:

Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

From Newton's law of universal gravitation we derive an expression:

       Fg  =  [tex]\frac{Gm_{1} m_{2} }{r^{2} }[/tex]

G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

Now insert the parameters and solve;

    Fg  = 6.67 x 10⁻¹¹ x [tex]\frac{100 x 300}{2^{2} }[/tex]    = 5 x 10⁻⁷N

Answer:

the blue one

Explanation:

Answer:

By teh way is isn't it question of law or science and the picture is of what electric light or not I have read it so I was asking isn't is question of science

Answer:

Explanation:

while it is still falling it is turned into kinetic energy. While on the floor it turns onto potential energy. Remember energy cannot be created or the destroyed so the other ones are automatically out.

Answer:

gap junctions,tight junctions,and desmosomes

Answer:

This is an example of Inelastic colission

Explanation:

Step one:

given:

mass of moose m1 = 620 kg

mass of train m2= 10,000kg

Initial velocity of moose  u1= 0 m/s

Initial velocity of train v1 = 10m/s

combined velocity of the system is given as v

Applying the conservation of momentum equation we have

m1u1+ m2u1= (m1+m2)V

substitutting we have

620*0+10000*10= (620+10000)V

100000= 10620V

divide both sides by 10620

V = 100000/10620

V=9.41m/s

The velocity of the moose after impact is 9.41m/s

Answer:

In picture one because string telephones are best heard when there is more tension on the string

Explanation:

Answer:

3 m³ will occupy more volume.

Explanation:

We need to tell which occupies more volume : 3m³ or 100 ft³.

To compare, we first make the units same.

We know that,

1 m = 3.28 feet

1 m³ = 35.31 foot³

3 m³ = 105.94 foot³

Now, we can compare 105.94 foot³ and 100 ft³. It can be clear that 3 m³ will occupies more volume.

Answer:

Angular velocity of merry-go-round is πk - 1 at t= T

Explanation:

From the question it is given that

[tex]\theta(t) = \pi k(t+k_e-\frac{t}{k} )[/tex] ..........................(1)

since mathematically, angular velocity is defined as

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] ........................(2)

on substituing the value of θ(t) from equation 1 in equation (2) we get

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] = [tex]\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}[/tex]   ............................(3)

on differentiating equation (3) with respect to time we get

ω(t) = πk(1 -[tex]\frac{1}{k}[/tex]) = πk - 1 angular velocity of merry-go-round

Therefore, angular velocity of merry-go-round is πk - 1 at t= T

The angular velocity of the merry-go-round at t=T is πk - 1. The pace at which angular displacement changes is defined as angular velocity.

The rate of change of angular displacement is defined as angular velocity.

It is stated as follows:

ω = θ t

The equation for the angular displacement is given as;

[tex]\theta (t) = \pi k (t+k_e-\frac{t}{k} )[/tex]

The angular velocity is found by the differentiation of the angular displacement.

[tex]\rm \omega(t) = \frac{d(\theta)}{dt} \\\\ \rm \omega(t) = \frac{d (\pi k (t+k_e-\frac{t}{k} )}{dt} \\\\ \rm \omega(t) = \pi k-1[/tex]

Hence the angular velocity of the merry-go-round at t=T is πk - 1.

To learn more about the angular speed refer to the link;

https://brainly.com/question/9684874

Answer:

  v_{f} = -0.693 m / s

Explanation:

The acceleration of the runner can be obtained from Newton's second law

       F = m a

        a = F / m

the bold are vectors, therefore the acceleration throughout the journey varies as the force has variations.

For the part of finding the velocity of the body we can use the relationship between the momentum and the variation of the momentum

            I = Δp

           ∫ F Δt = m [tex]v_{f}[/tex] - m v₀

           int F dt = m (v_{f}-v₀)

1) To find the change in velocity we must find the area under the curve of the graph, this can be done analytically if we know the functional of the curve or approximate it by intervals

a) between 0 <t <0.20 s

v) between 0.20 <t <0.30 s

a reasonable curve shape can be a Gaussian.

2) If we do not have the form of the cure, we can perform a graphical integration to find the area under the curve, we can do this by dividing the curve into small rectangles, finding the area of ​​each one and adding them.

3) Another even more approximate way is to create an average force in each interval and find the area of ​​this force, the average force is the average value of the force in the interval, let's use this method in the exercise

a) first interval 0 <t <0.20 Average force [tex]F_{mean}[/tex] = 300 N

    area = F_{mean} Δt

    area = I = 300 0.20

    I = 60 N s

the speed change is

    I = m Δv

    Δv = I / m

    Δv = 60/65

    Δv = 0.923 m / s

If we assume that the runner starts from rest, his final velocity is v = 0.923 m / s in the direction of the force.

b) second interval 0.2 <t <0.30s average force F_mean = 150 N

    area = I = 150 (0.30 - 0.20)

    I = 15 N s

the speed change is

    Δv = 15/65

    Δv = 0.23 m / s

Note that in this case the initial speed is not zero and since the two impulses are in the opposite direction the speed decreases

        [tex]v_{f}[/tex]= -0.923 + 0.23

        v_{f} = -0.693 m / s

Answer: Gravitational Forces

Explanation:

Answer:

0J

Explanation:

work done= force x distance in direction of force.

since distance=0, work done=0

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO  generally: (left or down) would be considered negative in an equation. And the other person  (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

0 is the number of equilibrium.

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

Answer:

[tex]v_2 = 6.406 / cos25.29[/tex]

Explanation:

From the question we are told that

Speed of snow ball 7.6m/s

Speed of chunk 9.3m/s at [tex]19 \textdegree[/tex]

Generally the equation for the conservation  of momentum is mathematically given as

Let the snow ball be b

and the chunk b/2

According to conservation of momentum we have

        [tex]b_u = (b /2) v_1 cos\theta_1 + (b / 2) v_2cos\theta_2[/tex]

        [tex]v_2 cos\theta_2 = 2 * 7.6 -9.3 * cos_19[/tex]

        [tex]v_2 cos\theta_2 = 15.2 -8.793 = 6.406[/tex]    

Therefore

          [tex](b/2)v_1 sin\theta_1 = (b/2) v_2sin\theta_2[/tex]

          [tex]v_2 sin\theta_2 = 9.3 * sin_19[/tex]

          [tex]v_2 sin\theta_2 = 3.027[/tex]

Mathematically From  above equations

        [tex]tan\theta_2 = 0.4726[/tex]

        [tex]\theta_2 = 25.29 \textdegree[/tex]

Therefore the speed and direction of second chunk

        [tex]v_2 = 6.406 / cos25.29[/tex]

        [tex]v2 = 7.0855m/s[/tex]

Given that,

Mass of Mr. Brooks = 100 kg

A chair lift up to 800 m to the top of the mountain.

To find,

How fast will he be going when he get back to the bottom of the mountain.

Solution,

Let v is the speed when he get back to the bottom of the mountain. Using the conservation of energy to find it.

[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 800} \\\\v=125.21\ m/s[/tex]

So, the required speed is 125.21 m/s.

Answer:

F = Changing momentum / time

Explanation:

change in momentum = final momentum - initial momentum

Answer:

50km i think but i am not very sure

Answer:

65N

Explanation:

Answer:

65 N

Explanation:

i just did the assignment on edge 2021

Answer:

450

Explanation:

Given,

Mass= 100kg

Velocity= 3 m/s

Kinetic Energy= ?

Kinetic Energy= 1/2 mv^2

= 1/2× 100× 3^2

= 1/2× 900

= 450.

HOPE IT HELPED :)

Answer:

Power of 5 watts is used

Explanation:

Mechanical Work and Power

Mechanical work is the amount of energy transferred by a force.

Being F the force vector and s the displacement vector, the work is calculated as:

[tex]W=\vec F\cdot \vec s[/tex]

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

[tex]\displaystyle P=\frac {W}{t}[/tex]

Where W is the work and t is the time.

It's required to calculate the power used in t=30 seconds when W=150 Joules of work are completed. Substitute in the formula:

[tex]\displaystyle P=\frac {150}{30}[/tex]

P = 5 Watt

Power of 5 watts is used

Answer:

a = 7.6\ mph/s

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

[tex]v_f=v_o+at[/tex]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

[tex]\displaystyle a=\frac{60 \ mph-0}{7.9}[/tex]

a = 7.6\ mph/s

Answer:

Looks like  streak color.

Explanation:

Answer:

[tex]\mathrm{a.\:}123\:\mathrm{kg\cdot m/s},\\\mathrm{b.\:}95\:\mathrm{kg\cdot m/s},\\\mathrm{c.\:}28\:\mathrm{kg\cdot m/s}[/tex]

Explanation:

The momentum of an object is given by:

[tex]p=mv[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.

a)

The mass of the child and bike together is [tex]19+5.6=24.6[/tex] kilograms. Since they're moving at a velocity of 5.0 m/s, their momentum is:

[tex]p=24.6\cdot 5=\fbox{$123\:\mathrm{kg\cdot m/s}$}[/tex].

b)

The mass of the child is given as 19 kg. Since the child is on the bike moving at 5.0 m/s, it's implied the child is as well. Therefore, the momentum of the child is:

[tex]p=19\cdot 5=\fbox{$95\:\mathrm{kg\cdot m/s}$}[/tex].

c) The mass of the bike is given as 5.6 kg and it is moving at 5.0 m/s. Therefore, the momentum of the bike is:

[tex]p=5.6\cdot 5=\fbox{$28\:\mathrm{kg\cdot m/s}$}[/tex]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

[tex]v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}[/tex]......(I)

The orbital speed of satellite B,

[tex]v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}[/tex]......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}[/tex]

Put the value into the formula

[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}[/tex]

[tex]\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}[/tex]

Hence, The ratio of their orbital speeds are 5:4.

Answer: 25%

Explanation:

From the question, we are informed that the temperature of source and sink of cannot engine are 400K

and 300K respectively.

The efficiency will be calculated as:

= 1 - t/T

= 1 - 300/400

= 1 - 3/4

= 1/4

= 25%

Answer:

2.8 A

Explanation:

We know that;

Vrms = Vo/√2

Where;

Io = peak value of voltage = 200V

Hence;

Irms = 0.707 × 200

Irms = 141.4 V

Irms = Vrms/R = 141.4/50 = 2.8 A

Answer:

The equivalent resistance is 2 ohms.

Explanation:

let the first resistance = R₁ = 10 ohm

let the second resistance = R₂ = 10 ohm

let the third resistance = R₃ = 10 ohm

let the fourth resistance = R₄ = 10 ohm

let the fifth resistance = R₃ = 10 ohm

The equivalent resistance is calculated as;

[tex]\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \\\\\frac{1}{R_T} = \frac{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)}{R_1R_2R_3R_4R_5} \\\\R_T = \frac{R_1R_2R_3R_4R_5}{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)} \\\\R_T = \frac{(10^5)}{(10^4)+(10^4)+(10^4)+(10^4)+(10^4)} \\\\R_T = \frac{10^5}{5(10^4)} \\\\R_T = \frac{10}{5} \\\\R_T = 2 \ ohms[/tex]

Therefore, the equivalent resistance is 2 ohms.