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Answer:

9v

Explanation:

Answer:

29 m/s

Explanation:

Assuming the bridge and the platform are back to back,

Average speed = Total Distance/ Total Time

Avg S = 1,450m/50s = 29 m/s

The average speed of the train is 29 m/s

The average speed is defined as the ratio of total distance traveled and the total time taken.

    Total distance = 650 m + 800 m = 1450 m

     Total Time      = 20 s + 30 s = 50 s

    Average speed = total distance/total time

                               = 1450 / 50

                               = 29 m/s is the average speed of the train.

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Answer:

The motion of an object is accelerated when its speed increases.

Explanation:

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Hi there! :)

[tex]\large\boxed{17.32 m/s}[/tex]

Use the following equation in solving for kinetic energy:

KE = 1/2mv² where:

KE = kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

12,000 = 40v²

Divide both sides by 40:

12,000 / 40 = v²

300 = v²

Take the square root of both sides:

√300 = v

v ≈ 17.32 m/s

The initial velocity of the projectile is 19.8m/s

Explanation:

First, find time.

From our kinematics equations:

delta y = Vi•t + (1/2)at^2

rearrange,

t = sqrt[(2•delta y)/a]

t = sqrt[(2•20m)/9.8m/s^2]

t = 2.02s

Next, plug time into new kinematics equation to solve for the Vi in the x direction (horizontal)

delta x = Vi•t + (1/2)at^2

delta x = Vi•t

Rearrange:

Vi = delta x/t

Vi = 40m/2.02s

Vix = 19.8m/s

Answer:

c is trueeeeeeeeeeeeeee

The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.

The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.

Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.

Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.

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Answer:

D

Explanation:

because if the solvent is more than the solvent then we can't resolve it.

so our product will be suspended

Answer:

0.116 N/m

Explanation:

Since the net force acting on the ring must be greater than 2.00 × 10⁻² N, and the surface tension T = F/L where F = net force = 2.00 × 10⁻² N and L = circumference of ring = 2πr where r = radius of ring = 2.75 cm = 2.75 × 10⁻² m.

So, T = F/L

= F/2πr

= 2.00 × 10⁻² N ÷ 2π(2.75 × 10⁻² m)

= 1/2.75π  N/m

= 1/8.64 N/m

= 0.116 N/m

Answer:

[tex]1.554\times 10^{32}\ \text{kg}[/tex]

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

Radius of orbit is given by

[tex]R=\dfrac{vT}{2\pi}[/tex]

We have the relation

[tex]\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}[/tex]

The mass of each star is [tex]1.554\times 10^{32}\ \text{kg}[/tex]

Answer:

The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.

Explanation:

I hope this helps

Answer:

[tex]f = g \times \frac{m1 \times m2}{ {d}^{2} } [/tex]

[tex]f = 6.67 \times {10}^{ - 11} \times \frac{40 \times 40}{9} [/tex]

Answer:

E =  1.64 x 10⁶ V/m

Explanation:

The electric field in the region between the plates can be given by the following formula:

[tex]E = \frac{\Delta V}{d}[/tex]

where,

E = Electric Field = ?

ΔV = Poetential Difference across the plates = 23 KV = 23000 V

d = distance between plates = 1.4 cm = 0.014 m

Therefore, using these values in the equation, we get:

[tex]E = \frac{23000\ V}{0.014\ m}[/tex]

E =  1.64 x 10⁶ V/m

Answer:

v = 9.04 m / s

Explanation:

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

          W = Em_f - Em₀         (1)

Starting point. Lower slope

        Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

        Em_f = U = m g h

The work of rubbing

        W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

        cos θ = W_y / W

        sin θ = Wₓ / W

        W_y = W cos θ

        Wₓ = W sin θ

Y axis

        N - Wₓ = 0

        N = mg sin  θ

X axis

         fr = m a

the friction force has the expression

         fr = μ N

         fr = μ mg sin θ

we look for the job

         W = - μ mg sin θ  x

where x is the distance along the slope

we substitute in 1

         -μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

        tan 30 = h / x

        x = h / tan 30

we substitute

          -   [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²

we use  

          tan 30 = sin30 / cos30

          v² = 2g h + 2 μ g h cos 30

          v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]

let's calculate

          v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]

          v = 9.04 m / s

The charge on each ball is 6.1 ×10⁻⁸ C.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.  Force is a vector quantity and it has both magnitude and direction.

Let the balls have equal charge of q. Then, electrostatic force acting on each particle = √3 q²/r² dyne =  √3 q²/15² dyne =0.0076 q².

Weight of each ball = 13.3 ×980 dyne = 13034 dyne.

As the electrostatic force and the weight acts perpendicular to each other.

Hence, 0.1/√3 = 0.0076 q²/13034

q = 6.1 ×10⁻⁸ C

Hence, the charge on each ball is  6.1 ×10⁻⁸ C.

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Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

Answer:

making an observation

Explanation:

Answer:

World War 1 was caused by entangled alliances, nationalism, imperialism, and major

advancements in military technology. Does the Treaty of Versaille address those issues?

Explain your answer using facts. (5 points)

Answer:

The answer is "892.90 N"

Explanation:

Following are the solution to these question:

Calculating the vertical force of the summation that is equal to zero:

[tex]\to TL \cos 65 + TR \cos 80 -520 = 0\\\\\to 0.4226\ TL + 0.1736\ TR = 520\\[/tex]

Calculating the sum of horizontal forces that is equal to zero:

[tex]\to TL\sin 65 - TR \sin 80 = 0 \\\\\to 0.9063TL - 0.9846TR = 0\\\\\to TL = (\frac{0.9846}{0.9063})\ TR \ \ = 1.0866\ TR\\\\\to 0.4226(1.0866) \ TR +0.1736\ TR =520 \ N\\\\\to 0.6328 \ TR = 520 \\\\\to TR = 821.74 \ N \\\\\to TL = 1.0866 \times 821.74 = 892.90\ N[/tex]

Explanation:

Check out the Periodic Table attached here. The Carbon family is the vertical group starting with Carbon, Group 14. Nonmetals are in green, which element there is a nonmetal?

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

The specific gravity of the rock in the given water volume is 0.2.

The given parameters;

The specific gravity of the rock is calculated as follows;

[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]

Thus, the specific gravity of the rock in the given water volume is 0.2.

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Answer:

option 5

Explanation:

because all u do is have to add them up

This problem involved half projectile.

initial velocity, vo = 20 m/s

time of flight, t = 7 s

(a) Simply use the formula to get the height, h:

h = vo*t - (1/2)gt^2

(b) To get the final vertical velocity or terminal velocity (vf), use the formula:

(vf)^2 - (vo)^2 = 2gh

(c) Use the formula find the horizontal distance traveled, R:

R = vo * cos(θ) * t

But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have

R = vo * t

Hope this helps~ `u`

Jai

Answer:

3.39 rad / s.

Explanation:

Given data:

mass of disk = 4.50 Kg

radius of wheel = 0.650 m

mass of the clay = 0.870 kg

The moment of inertial of the wheel = I = 4.5 kg x ( 0.65 m )2 / 2 = 0.95 kg . m2.

Now, applying the principle of angular momentum conservation :

Iω_i = ( I + mr2 )ω_f.

where ω_i = initial angular speed= 4.70 rad/s, ω_f = final angular speed

Hence, ω_f = Iω_i / ( I + mr2 )

= ( 0.95 kg . m2 x 4.7 rad / s ) / [ 0.95 kg . m2 + 0.87 kg x ( 0.65 m )2 ]

= 3.39 rad / s.

Hence, correct answer is  : 3.39 rad / s.

Answer:

Drought

Explanation:

Answer:Air bags can leave you in even more injury, From the impact they give

You could end up with a broken nose,arm

concussion

Answer:

igneous = melted rocks formed by cooled magma

sedimentary is brken rocks, kayers with fossil...

metamophic is rocks formed by pressure and heat