Convert 580 g to kg.

Answer:

The net force should be of a magnitude of 64 N

Explanation:

We use Newton's second Law for this:

[tex]F_{net} = m\,*\,a[/tex]

which for our case gives:

[tex]F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N[/tex]

Answer:

D) Humans

Explanation:

Photosynthesis is a useless ability without some way of exposing yourself to as much of the Sun's energy as possible. That requires a large surface area, relative to their volume. Plants achieve that with large, horizontal, light-capturing surfaces – leaves.

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =        mass    

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

If A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, then the mass of the air in the classroom is 2322Kg.

Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water).

Given,

Height = 3 m

Width = 20 m

length= 30 m

Density of air = 1.29kg/m³

The  volume of the room = 3×20×30 m³

Volume V = 1800m³

By formula,

Density = Mass/Volume

1.29kg/m³ = Mass/1800m³

Mass of the air = 1.29×1800 = 2322 Kg

The mass of the air is classroom is 2322Kg.

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Answer:

the winds will make the bird stop

Explanation:

is basically 20 - 20

Answer:

Control groups let the one who is expermenting compare  the effect of the varibles in the expermental group.

Explanation:

Answer:

20.1 m/s

Explanation:

Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.

Given that the shuttle has a constant acceleration of 4.5 m/s2. 

The total distance to cover is:

Total distance = 40.9 + 3.9 = 44.8 m

Assuming you are starting from rest. Then initial velocity U = 0

Using the 3rd equation of motion to calculate the minimum velocity.

V^2 = U^2 + 2as

V^2 = 0 + 2 × 4.5 × 44.8

V^2 = 403.2

V = sqrt (403.2)

V = 20.1 m/s

Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s

Answer:

Pluto & Mercury

Explanation:

Pluto's eccentricity is 0.248

Mercury's eccentricity is 0.206

Answer:

experimental group

Explanation:

please mark me as a brainlist..

Answer:

Experemtial group

Explanation:

The placebo is just sugar pills tested is ones already done and control is a mix of the two

Answer:

[tex] \boxed{\sf Distance \ travelled = 12 \ m} [/tex]

Given:

Initial speed (u) = 0 m/s (Accelerates from rest)

Final speed (v) = 8 m/s

Time taken (t) = 3 seconds

To Find:

Distance travelled by cyclist (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

[tex] \boxed{ \bold{s = (\frac{v + u}{2} )t}}[/tex]

By substituting value of v, u & t in the equation we get:

[tex] \sf \implies s = ( \frac{8 + 0}{2} ) \times 3 \\ \\ \sf \implies s = \frac{8}{2} \times 3 \\ \\ \sf \implies s = 4 \times 3 \\ \\ \sf \implies s = 12 \: m[/tex]

[tex] \therefore[/tex]

Distance travelled by cyclist (s) = 12 m

Answer:

s(distance) =36m

Explanation:

u(initial velocity) =0 m/s

a =8 m/s^2

t=3s

s=ut+1/2at^2

s=1/2(8)(3)^2

s=1/2(8)(3)(3)

s=4(9)

s=36m

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

Answer:

Straight line in the direction of the tangential velocity the ball had at the moment the string broke

Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

Answer:

B

Explanation:

just took the test :)

Explanation:

This problem is on units conversion

given the dimensions

3.12 ft in length,

0.0455 yd in width and

7.87 inches in height

we have to convert to cm first

3.12 ft to cm=

1 ft-------30.48cm

3.12--------x

x=95.09cm

0.0455 yd to cm

0.0455yd ft to cm=

1 yd-------91.44cm

0.0455--------x

x=4.16cm

7.87 in to cm

1 in-------2.54cm

7.87--------x

x=19.98cm

the volume is =95.09cm*4.16cm*19.98cm

volume= 7907.45cm^3

the volume in gallon is

1 cm^3------0.000264172gal

7907.45cm^3-------x

x=7907.45*0.000264172

x=2.088gallons

Answer:

Hello!

Sorry you haven't put up an image of your question! Without it we can't answer your question!

Explanation:

Maybe put up another one and it'll be answered!

:D

Answer:

ans is 120

Explanation:

I= 0.5 A

p= 60w

p=Iv

60/0.5

v= 120

Answer:

V=P/I

V=60/0.5

V=120

Explanaton:

the formula P=IV is converted to V=P/I to make V (voltage) the subject. then sub in the values to get your answer

note that potential difference is same as voltage

they get their energy from photosynthesis, so the answer would be (B) the sun

Answer:

(a)  the frequency of the lowest note the piccolo can sound is 571.7 Hz

(b) the distance between adjacent antinodes is 5.72 cm

Explanation:

(a)

Given;

length of piccolo, L = 30 cm = 0.3 m

the speed of sound in air is 343 m/s

The wavelength of a pipe open at both ends, for the first harmonic is given;

L = A → N + N → A

L = λ / 4 +  λ / 4

L = λ / 2

λ = 2L

λ = 2 x 0.3 = 0.6 m

The fundamental frequency (lowest frequency) is given  by;

f₀ = v / λ

f₀ = (343 / 0.6)

f₀ = 571.7 Hz

(b)

Given;

highest note, f = 3000 Hz

the distance between adjacent antinodes is given by;

[tex]d = \frac{v}{2f}\\\\ d = \frac{343}{2*3000}\\\\ d = 0.0572 \ m\\\\d = 5.72 \ cm[/tex]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

Answer:

The tensions in [tex]T_{BC}[/tex] is approximately 4,934.2 lb and the tension in [tex]T_{BD}[/tex] is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = [tex]T_{BC}[/tex]

The tension in rope segment BD = [tex]T_{BD}[/tex]

The tension in rope segment AB = [tex]T_{AB}[/tex] = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = [tex]T_{AB}[/tex] = 1200 lb

[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

[tex]T_{BC}[/tex] × sin(26.0°) + [tex]T_{BD}[/tex] × sin(21.0°) = 0...........................(2)

Which gives;

[tex]T_{BC}[/tex] × sin(26.0°) = - [tex]T_{BD}[/tex] × sin(21.0°)

[tex]T_{BC}[/tex] = - [tex]T_{BD}[/tex] × sin(21.0°)/(sin(26.0°))  ≈ - [tex]T_{BD}[/tex] × 0.8175

Substituting the value of, [tex]T_{BC}[/tex], in equation (1), gives;

- [tex]T_{BD}[/tex] × 0.8175 × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb

- [tex]T_{BD}[/tex] × 0.7348  + [tex]T_{BD}[/tex] ×0.9336 = 1200 lb

[tex]T_{BD}[/tex] ×0.1988 = 1200 lb

[tex]T_{BD}[/tex] ≈ 1200 lb/0.1988 = 6,035.6938 lb

[tex]T_{BD}[/tex] ≈ 6,035.6938 lb

[tex]T_{BC}[/tex] ≈ - [tex]T_{BD}[/tex] × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

[tex]T_{BC}[/tex] ≈ -4934.1733 lb

From which we have;

The tensions in [tex]T_{BC}[/tex] ≈ -4934.2 lb and  [tex]T_{BD}[/tex] ≈ 6,035.7 lb.

The answer is OPTION B : 73.7 N

Answer:

B

Explanation:

Answer:  Heterogenous mixture

Explanation:

A heterogeneous mixture in which the components of the mixture are not evenly mixed or uniform, allowing one to identify the different constituents of each components and enable the mixture to be separated physically.

Examples of heterogeneous mixtures includes

-- sand and nails

---rice and beans

--- water and oil

Another type of mixture is the homogeneous mixture in which all the components are evenly mixed causing that each components cannot be visible with the eye and therefore separated chemically.

Answer:

Acceleration /Speed

Explanation:

An objects Velocity can be determined by acceleration,

Please pay attention in your middle school class, speed and velocity quiz.

Answer:

In the - j direction, that is negative of the y-axis

Explanation:

As typed in the question, the position of the object is given by the expression in three component ( i, j, k) form:

r (t) = 5  i  - (t + 1 )  j  + t^3  k

and since the velocity is the derivative of position with respect to time, by doing the derivative of this expression we get:

v(t) = 0  i  -  1  j   +3 t^2  k

which for the initial velocity requested (that is at time zero) we have:

v(t) = 0  i  -  1  j   +3 (0)^2  k = = 1  j

Then the direction of the initial velocity is entirely in the direction of the j versor, that is pointing to the negative of the y-axis.

Answer:

D. The volume should be divided by temperature on each side of the equation.

Explanation:

Charles's Law is  [tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }[/tex].

Answer:

The volume should be divided by temperature on each side of the equation.

Explanation:

Answer:

The answer is tiny organisms known as cyanobacteria, or blue-green algae. These microbes conduct photosynthesis: using sunshine, water and carbon dioxide to produce carbohydrates and, yes, oxygen. In fact, all the plants on Earth incorporate symbiotic cyanobacteria (known as chloroplasts) to do their photosynthesis for them down to this day.

Explanation:

Answer:

high frequent sound

Explanation:

because if its low than its slower.

Explanation:

100cm=1m

5 cm= x

cross multiple

x=5/100

=0.05m

There are 0.05 meters in 5 cm. The correct option among the following is option (B).

In the metric system, a centimeter (cm) is a unit of length. It is one centimeter (1 cm = 0.01 m) in size. Small distances or measurements, such as the length, height, or width of items, are frequently measured in centimeters. Due to its practical size for many routine measurements, it is also frequently employed in scientific and mathematical computations.

In the International System of Units (SI), the meter serves as the base unit and is frequently used to measure lengths, heights, and other measurements. It serves as the basic building block for other metric units like centimeters (one meter equals one hundred centimeters) and kilometers (one kilometer equals one thousand meters).

So, to convert 5.0 cm to meters:

5.0 cm ÷ 100 = 0.05 m

Hence, 5.0 cm is equal to 0.05 meters. The correct option is (B).

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This question is incomplete, the complete question is;

The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor

Assume 12ft of elevation per floor

Answer: 48.68 psig

Explanation:

First  we calculate the elevation of the building

hb = 27 story * 12ft per floor/story

hb =  324 ft

given that the head lost in the vertical riser hL = 40 ft

now the delivery head required in the riser on he 27th floor;

hd = 8 psig *  (2.31 ft / 1 psig)

hd = 18.46 ft

Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume

hp = (hb + hL + hd) - hs

hs = hb + hL + hd - hp

where hp is the head developed by the pump (270 ft)

hb is the elevation of the 27th floor of the building ( 324 ft)

hL is the head lost in the vertical riser ( 40 ft)

hd is the head required to exist in the riser on the 27th floor (18.46 ft)

so we substitute

hs = 324 ft + 40 ft + 18.46 ft - 270 ft

hs = 112.46

so 112.46ft * (1 psig / 2.31 ft)

= 48.68 psig

Answer:

[tex]x=119m[/tex]

Explanation:

Hello,

In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:

[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]

In such a way, we can compute the displacement via:

[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]

Best regards.