PLEASE HELP ASAP
I need help understanding this question, form my guess I’m thinking it’s C but I’m not to sure

Answer: push force gravity tension and reverse force

Explanation: sense they are pushing the box there is a push force gravity because this is likely on earth tension because it is the reverse of gravity and the reverse force because you have to have the reverse of push

Answer:

Option C, increases and decreases

Explanation:

When an object making noise approaches you, the wave frequency increases leading to a higher pitch. Conversely, when it moves away from you or retreats, the wave frequency decreases leading to a lower pitch. This can be observed in ambulance sirens.

Answer:

The altitude of the satellite above the surface of the Earth is 1,528.48 km.

F=rmv2=R2GmMrv2=R2GMr=GMv2R2r=(6.67259×10−11)×(5.98×1024)(6900)2×(6,370,000)2r=4,841,516.6 mr=4,841.52 km

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:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

The charge moved by this electric force is [tex]5.4 \times 10^{-5} \; Coulombs[/tex].

Given the following data:

To determine the charge:

Mathematically, the work done by an electric force in moving a charge from one point to another is given by the formula:

[tex]W = PQ[/tex]

Where:

Making Q the subject of formula, we have:

[tex]Q =\frac{W}{P}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Q =\frac{2.70 \times 10^{-3}}{50}\\\\Q =5.4 \times 10^{-5} \; Coulombs[/tex]

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The answer is:

That they are switched on

Answer:

a

Explanation:

i took the test 100%

Answer: The answer is A) 49 N(Newtons).

Answer:

you forgot to attach the image

Answer:

V= A ω      maximum KE of object in SHM

V2 / V1 = .958     ratio of amplitudes since ω is constant

KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2

KE2 / KE1 = .958^2 = .918

So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle

Answer:

Explanation:

Not in Genesis. The proposed sacrifice of Isaac was all Abraham's doing. Sarah was not present and could not, therefore offer any input. It might have been covered in other Jewish writings, but it is not recorded in Genesis.

Since the book most widely used by Everyone is the Bible, I would answer false.

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:

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:

c. Impulse

Explanation:

Answer:

[tex]v^2=\frac{T}{M}[/tex]

where v = transverse wave velocity, T = tension in the string, M = mass per unit length.

Explanation:

Dimensional analysis is where you just look at the units and see how they fit within each other.

In this case, all relationships are made using the MLT comparisons, where M stands for Mass, L stands for Length, and T stands for time.

For example, for velocity, we know the SI unit for velocity is [ms⁻¹] which is L¹T⁻¹, we can do the same thing for tension = [N = Kgms⁻²] = M¹L¹S⁻², and for the mass per unit length which we can think of as [Kgm⁻¹] = M¹L⁻¹.

If you play around a little with the powers, you can find a relationship:

[tex]v^2=\frac{T}{M}[/tex]

since:

[L¹T⁻¹]² = L²T⁻² = M¹L¹T⁻² ÷ M¹L⁻¹ = M¹⁻¹L¹⁻⁽⁻¹⁾T⁻² = M⁰L²T⁻² = L²T⁻²

Answer:

The astronomical unit is used primarily for measuring distances within the Solar System or around other stars.

Answer:

I think some drawbacks are that since there are no solid walls meaning it is weak and if murphy's law is in place, the water will destroy the substance. Tsunami waves also happen very quickly so even if the water can travel thru the substance quickly, it probably won't be quick enough. This design could help if the wave is smaller because less destruction would occur.

Explanation:

yeah

Answer:

net force would be 50 N right

Explanation:

Yes, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.

According the principle of conservation of energy, the total kinetic energy is equal to the total potential energy.

[tex]P.E _i + K.E_i = P.E_f + K.E_f \\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\g(0) + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\\frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\v_i^2 = 2gh_f + v_f^2\\\\v_i^2 - 2gh_f = v_f^2\\\\v_f = \sqrt{v_i^2 - 2gh_f}[/tex]

where;

Thus, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.

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Answer:

strong nuclear force.

Explanation:

hope this helps you!!

Answer:False

Explanation:

the intensity of the sound wave decreases with increasing distance from the source.

The conventions for the current allow to find the result for which is the direction of the current the is circuit is:

Current is the movement of electrons in a circuit per unit time.

In an electrical circuit the electrons that have a negative charge leave the negative electrode towards the positive electrode, by convention the current goes in the opposite direction from the positive electrode towards the negative electrode.

When analyzing the given circuit, the current leaves the positive pole at the top of the battery and moves through the blue wire, reaches the light bulb and moves the red wire and reaches the negative pole of the battery.

In conclusion using the conventions for the current we can find the result for which is the direction of the current the is circuit is:

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a) Sketches of all possible pv-diagrams for the cycle are attached below

b) The work W[tex]_{ac}[/tex] for the process Ca is : 2462.8  J

Given data :

Amount of heat flowing out = 800 J

Ta = 200 K

Tb = 300 K

R = 800

B) Determine the work W for the process Ca

Wₐs = -pdv

       = - [ pVb - pVa ] ---- ( 1 )

note : pVb = nRTb ,  pVa = nRTa

Equation ( 1 ) becomes

= -nR [ Tb - Ta ]

= - 2(8.314 ) [ 300 - 200 ]

= - 1662.87

given that W[tex]_{bs}[/tex] = 0 which is isochonic

dv = 0 ( cyclic process ) = d∅ - dw

∴ 0 = 800 - ( Wₐs  +  W[tex]_{ac}[/tex] )

Therefore : W[tex]_{ac}[/tex] = 800 + 1662.8 = 2462.8  J

Hence we can conclude that the work W for the process Ca = 2462.8  J

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