a rocket ship is moving through space at 1000 m/s. It accelerates in the same direction at 4m/s/s. What is its speed after 100 seconds

The volumetric flow rate if friction is negligible = 0.00197 m³/sec

Given data :

P₁ = 7 atm  =  709.275 kPa

P₂ = 1.5 atm = 151.988 kPa

d₁  = 0.74 cm = 0.0074 m

d₂ = 1.20 cm  = 0.0120 m

h = 82 m

density of water ( p ) = 1000 kJ/m³

Estimate the volumetric flow rate

given that friction losses are negligible

V₁² - V₂² = [tex]\frac{2(P_{2} - P_{1}) }{p} + 2g(h )[/tex]

             = 2( 151.988 - 709.275 )  + 2 * 9.81 * 82

             = 2 ( - 557.287 ) + 1608.84

             = - 1114.574 + 1608.84  =  494.266

∴ V₁² - V₂² =  494.266 ----- ( 1 )

Also given that:

volumetric flow rate ( Q ) = A₁V₁ = A₂V₂

                                          = π / 4 ( 0.0074 )² * V₁   =   π / 4 ( 0.0120 )² * V₂

                                          = 0.000043 V₁ = 0.000113 V₂

∴ V₁ = 2.63 V₂

Back to equation ( 1 )

( 2.63 v₂ )² - V₂² = 494.266

V₂² = ( 494.266 ) / 1.63

∴ V₂ = √303.23 = 17.41  m/sec

Therefore the volumetric flow rate ( Q )

Q = A₂V₂

   =  π / 4  * ( 0.0120 )² * 17.41

   = 0.00197 m³/sec

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The vertical component of the applied force on the sled is 192.84 N.

The given parameters:

The vertical component of the 300 N force is calculated as follows;

[tex]F_y= F \times sin(\theta)\\\\F_y = 300 \times sin(40)\\\\F_y = 192.84 \ N[/tex]

Thus, the vertical component of the applied force on the sled is 192.84 N.

Your question is not complete, the missing part of the question is below:

The vertical component of the 300 N force is approximately?

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Answer:

Explanation:

if you are pulling in the positive direction and Upward is also positive.

F = -0.96i + 0.58(9.8)j

F = -0.96i + 5.7j

F = √(-0.96² + 5.7²) = 5.8 N

You are pulling a sled with constant velocity by using a rope then the surface force the ground exerts on the sled will be 5.8 N.

The pulling force conveyed axially by a string, cable, chain, or another analogous object, or by either end of a rod, truss member, or another such three-dimensional object, is referred to as tension in physics. Another way to think of tension is as the action-reaction pair of forces acting at each end of the previous elements. Perhaps tension is the polar opposite of compression.

A restoring force may cause what is also referred to as tension when atoms or molecules are torn apart from one another at the atomic level and build up potential energy.

If you are moving upward and pushing in a positive way,

F = -0.96 i + 0.58(9.8) j

F = -0.96 i + 5.7 j

F = [tex]\sqrt{(-0.96^2 + 5.7^2)}[/tex]

F= 5.8 N

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Answer:

net force would be 50 N right

Explanation:

Those organic structures that do not seem to play any important biological function in the organism that possesses them are known as vestigial structures.

A vestigial structure is one that has atrophied or lost its original function throughout the course of evolution.

Therefore, we can conclude that vestigial structures are structures that some organisms still conserve as an inheritance from their ancestors but that, for some reason or another, are no longer useful for the functioning of the same.

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42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]

[tex]\Rightarrow mg\sin{\theta} = F[/tex]

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get

[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]

or

[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]

[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]

[tex]\;\;\;=42.9°[/tex]

Answer:

The net force on the hanging mass is thus

2 T Mg     18 6.6 N 11.4 N

, enough to accelerate it upward at

17 m/s2

. The angular acceleration of the pulley is thus

   

2 2

Explanation:

The tension in the rope on both sides of the pulley 20.21N.

To solve this problem, we'll need to consider the forces acting on both sides of the pulley and apply Newton's second law of motion. Let's break it down step by step:

(a) Is the tension in the rope the same on both sides of the pulley?

No, the tension in the rope is not the same on both sides of the pulley. The side with the greater mass attached will experience a greater tension in the rope.

(b) Find the tension in the rope on both sides of the pulley:

Calculate the gravitational force on each mass:

Gravitational force on the pulley: Fpulley = mpulley * g, where mpulley is the mass of the pulley and g is the acceleration due to gravity (approximately 9.8 m/s²).

Gravitational force on the hanging mass: Fhanging = mhanging * g, where m_hanging is the mass of the hanging mass.

Calculate the net force on the pulley:

Net force on the pulley is the difference between the tension forces on either side: Fnet = TLeft - Tright.

Apply Newton's second law to the pulley:

For the pulley, Fnet = mpulley * a, where a is the acceleration of the pulley. Since the pulley is assumed to be massless, we can use the relationship a = α * r, where α is the angular acceleration and r is the radius of the pulley.

Use the relationship between linear acceleration and angular acceleration:

α = a / r.

Equate the torque due to the tension to the moment of inertia times the angular acceleration:

τ = I * α, where τ is the torque, I is the moment of inertia of the pulley, and α is the angular acceleration.

Substitute the expression for α and solve for the net tension:

Tnet = (τ / r) = (I * α) / r = (1/2 * mpulley * r² * α) / r = (1/2 * mpulley * r * a).

Now, substitute the expression for a from step 3 and solve for the net tension:

Tnet = (1/2 * mpulley * r * α) = (1/2 * mpulley * r * (a / r)) = (1/2 * mpulley * a).

Substitute the expression for a from step 2 (Fnet = mpulley * a) and solve for the net tension:

Tnet = (1/2 * Fnet).

Now, you can find the tensions on each side of the pulley:

Tleft = Tnet + Fhanging

Tright = Tnet - Fpulle

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Answer:

Explanation:

Do you mean 7 x 10⁻⁵² seconds? Who can even measure that small a time period?

If that time is incorrect, just redo the math with the correct time to get the force.

An impulse will cause a change of momentum

FΔt = mΔv

F = mΔv/Δt

F = 2(0 - 6)) / (7 x 10⁻⁵² - 0)

F = - 1.7 x 10⁵² N

The magnitude of that force would pretty much crush any rock it hit.

Answer:

0.47 J

Explanation:

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

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Answer:

Explanation:

I will ASSUME that the cylinder floats with the axis through the center of its flat ends is perpendicular to the water surface.

The cylinder will displace it's mass in water.

water density has a value of 1000 kg/m³

if the depth of the bottom is "h" in meters

1.80 kg = π(0.018² m²)(h m)(1000) kg/m³

h = 1.768388256...

or about 1.77 m

Answer:

Explanation:

A) 215 + 45.0 = 260 km/h West

B) 215 - 45.0 = 170 km/h West

C) √(215² + 45.0²) = 219.6588... 220 km/h

θ = arctan(45.0/215) = 11.8214... 11.8° S of W

The airplane's ground velocity is westward at 170 km/h regardless of wind direction. The south wind does not affect the plane's westward velocity in component C.

Vector addition will determine each scenario's plane velocity relative to the ground. We'll simplify by assuming a consistent horizontal breeze.

Let's divide the airplane's velocity and wind velocity into westward (negative) and eastward (positive) components.

Data: Plane airspeed = 215 km/h.

45 km/h wind.

A. West wind:

The airplane's westward (negative) ground velocity:

West aeroplane velocity = -215 km/h

-45 km/h west wind.

Find Vg when the wind blows west:

Vg (west) = Aeroplane velocity + Wind velocity (west)

Vg(west) = -215 km/h - (-45 km/h)

Vg (west)=-170 km/h

B. East wind: 2. Westward (negative) aeroplane velocity relative to the ground:

West aeroplane velocity = -215 km/h

Wind velocity (east) = 45 km/h (because the wind is blowing eastward)

Find Vg when the wind blows east:

Vg (west) = Aeroplane velocity (west) + Wind velocity (east).

Vg (west) = -215 + 45 km/h

Vg (west)=-170 km/h

C. Southerly wind: 3. Westward (negative) aeroplane velocity:

West aeroplane velocity = -215 km/h

Wind velocity (south) = 0 km/h (no effect since wind is not moving westward)

Find Vg when the wind blows south:

Vg (west) = Aeroplane velocity (west) + Wind velocity (south).

Vg (west) = -215 + 0 km/h.

Vg (west)=-215 km/h

All resultant velocities are westward and 170 km/h.

Part C's plane's westward velocity is unaffected by the south wind. The plane's ground velocity remains -215 km/h (westward). Since the heading is westward, no angle is needed.

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Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

To solve this we must be knowing each and every concept related to  centripetal force and its calculations. Therefore, the centripetal force acting on her is 3430 N.

The term centripetal relates to a propensity to gravitate toward the center. Centripetal refers to moving in the direction of the center. The force that maintains an item moving in a circular direction and helps it stay on the path is referred to as centripetal force.

Furthermore, centrifugal force is indeed the tendency of things to deviate from a circular route and fly in a straight line. People frequently confuse centripetal force with centrifugal force.

Mathematically,

F = m a

  = 70 acceleration

  = 70 × 5 × 9.81

 = 3430 N

Therefore, the centripetal force acting on her is 3430 N.

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Answer:

goalindia

goalindiaGiven:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kg

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 W

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s

goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s- So, the velocity of the car will be 0.61❤

Answer:

I am pretty sure it is called energy.

Answer:

Explanation:

Near the earth's surface where gravity is approximately 10 m/s² downward

v = u + at

v = 5 + (-10)(1) = -5 m/s

so it has the same speed but in the opposite (downward) direction.

The instantaneous velocity of the object is -4.8 m/s and is directed downwards.

Instantaneous velocity is defined as the rate of change of position for a given interval of time. And it also represents the direction.

Here,

The object is freely falling.

So, the acceleration of the object is the acceleration due to gravity.

a = g = 9.8 m/s²

Initial velocity of the object, u = 5 m/s  (upward velocity)

Time given, t = 1

Using the first equation of motion,

v = u + at

v = 5 + (-9.8) x 1  

( negative sign is due to it is directed downwards which is opposite to u)

v = -4.8 m/s

That means the instantaneous velocity is directed downwards.

Hence,

The instantaneous velocity of the object is -4.8 m/s and is directed downwards.

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Answer:

is there any questions

Answer:

What is your question???

Explanation:

Answer:

inertia

Explanation:

the mass of the box will want to continue on its straight line motion.

As centripetal force on the wheel treads causes the truck to turn, frictional force between the box and truck bed is insufficient to keep the box turning at the same rate that the truck turns, so relative velocity occurs between box and bed until the centripetal force is increased by contact with the truck side wall farthest from the center of curvature.

Answer:

Decant it.

Explanation:

Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.

Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

Answer:

The main condition necessary for a controlled thermonuclear fusion:

The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.

Answer:

Explanation:

Without the diagram, we have to bracket our answers.

Total energy is constant, so we eliminate options A and B.

C is correct if position 1 is closer to the star than position 2

D is correct if position 2 is closer to the star than position 1

Answer:

Water speed Plays a key role In determining The shape of it

Answer:

The movement of Tectonic plates.

Explanation:

Answer:

Explanation:

Only force in the direction of motion does work

The force acts in the x direction

The x change of position was 6 m

W = Fd = 12(6) = 72 J

The work done by the given force during this displacement is determined as 120 J.

The work done on the object is calculated as follows;

W = Fd

where;

d = √(6² + 8²)

d = 10 m

W = 12 x 10

W = 120 J

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Answer:

Hey Friend...

Explanation:

This is ur answer.....

Hope it helps!

Brainliest pls!

Follow me! :)

Answer:

4

Explanation:

element is a substance that cannot be broken down into any other substance which means every element is made up of its own type of atom

Answer:

Explanation:

Itz Deepika Here to help ~ .

Note:- If you have more questions then let me know !!

Answer:

Explanation:distance-time =speed

a,V =s/t

V=2m/2s

V=1m/s

b,v=s/t

V=80m/40s

V= 2m/s

The average speed is 2m/s