For a wave, the _____ the amplitude, the _____ energy the wave carries.
Multiple choice question.

A)
larger, more

B)
smaller, more

C)
larger, less

D)
smaller, same

This question involves the concepts of Gravitational Force and Weight force.

The value of acceleration is "b. 7.32 m/s²".

At the given height the weight of the rocket must be equal to the gravitational force between rocket and the Earth:

[tex]W=F_G\\mg=\frac{GmM}{R^2}\\\\g=\frac{GM}{R^2}[/tex]

where,

g = acceleration = ?

G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = mass of earth = 5.97 x 10²⁴ kg

R = Radius of Earth + Height = 6.378 x 10⁶ m + 1 x 10⁶ m = 7.378 x 10⁶ m

Therefore,

[tex]g=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{(7.378\ x\ 10^6\ m)^2}[/tex]

g = 7.32 m/s²

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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:

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:

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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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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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:

bro what.

Explanation:

... is this an actual physics question?

Answer:

over 50% power ability

Explanation:

I believe it's half and half chance that The Offspring or the child will have that trait

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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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:

I am pretty sure it is called energy.

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

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.

The speed of the cannon after firing is 1.2 m/s

This can be solved using the law of conservation of momentum.

From the law of conservation of momentum,

⇒ The momentum of the projectile is equal to the momentum of the cannon.

⇒ Where :

⇒ make V the subject of the equation

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, The speed of the cannon after firing is 1.2 m/s

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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:

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:

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:

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

Answer:

Explanation:

Itz Deepika Here to help ~ .

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

Answer:

The movement of Tectonic plates.

Explanation:

Answer:

900N, charges repel

Explanation:

F = KQq/d²

K = 9 × 10^9Nm²/C²

Q= 10mC = 10 × 10^-3C

q = 1mc = 1 × 10^-3C

d = 10m

F = ?

Force = (9 × 10^9 × 10 × 10^-3 × 1 × 10^-3)/10²

= 9 × 10²

= 900N

It will be an electrostatic force of repulsion since like charges(two positive charges) repel

The magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.

Coulomb's law states that whenever the two charged particles are separated by a particular distance then there will be a force of attraction or repulsion acts between the charge.

The formula of coulomb's force will be given by

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]K=9\times 10^9\ \frac{Nm^2}{C^2}[/tex]

[tex]q_1= 10\times 10^{-3}\ C[/tex]

[tex]q_2=1\times 10^{-3}\ C[/tex]

[tex]\rm Distance \ d=10 \ m[/tex]

Now the force will be calculated as

[tex]F=(9\times 10^9)\dfrac{(10\times 10^{-3})\times (1\times 10^{-3})}{10}[/tex]

[tex]F= 900\ N[/tex]

The charges are positive in nature so they will repel each other

Hence the magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.

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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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His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.

Answer:The comb and hair have become charged with opposite charges.

Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.

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:

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