The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room and compare your measured value to the accepted value.

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

Answer: 55 kmph

Explanation: Divide 110 by 2

Answer:

me...:(

Explanation:

Answer:

hello I'm online here thanks for the points (◔‿◔)

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.

It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex]    and   [tex]$\phi_{B,f} = 0$[/tex]

Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]

  [tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]

  = 30.27 V

Therefore, the emf generated is 30.27 V.

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

Explanation:

4. Alin sa mga sumusunod na awitin ang may tempong presto?

a. “Chua-ay”

c. “Akong Manok”

b. “Sitsiritsit

d. "Ili-ili Tulog Anay”

5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?

a. mabagal

c. mabilis na mabilis

b. mabilis at mabagal d. katamtamang bilis

6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?

a. descant

b. ostinato

c. tempo

d. texture

7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

[tex]F = V\times B[/tex]

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

Answer:

Explanation:

His Kinetic energy = 1/2 m v^2

v = 5 m/s

m = 75 kg

Ke = 1/2 75 * 5^2

Ke = 937.5 Joules

This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE

PE = m * g * h

m = 75

g = 9.81

h = ?

PE = 937.5

937.5 = 75 * 9.81 * h

937.5 = 735.75 * h

937.5/735.75 = h

h= 1.27 meters

Answer: 10 J

Explanation:

Trust ;)

Answer:

81.1 m/s

Explanation:

The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration

So, T - f = ma

T -  μmg  = ma

a = T/m - μg

susbstμituting the values of the varμiables into the equation, we have

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 2.67 m/s² - 0.98 m/s²

a = 1.69 m/s²

Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s

So, v = u + at

v = 0 m/s + 1.69 m/s² × 48 s

v = 0 m/s + 81.12 m/s

v = 81.12 m/s

v ≅ 81.1 m/s

So, Jason's top speed is 81.1 m/s

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

∴

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

Answer:

The centripetal acceleration is 6.95 m/s²

Explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²

Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

Answer:

4.408 [tex]\mathsf{M_{sun}}[/tex]

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]

combined mass (m₁+m₂)  = 4.408 [tex]\mathsf{M_{sun}}[/tex]

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

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Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

Use formula:

Replace and solve:

The tension in the rope is 98 Newtons.

Greetings.

There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options

Answer:

8760 N

Explanation:

think this is the right answer :)

Answer:

A. It must stay the same, but kinetic energy (KE) can be transformed to PE and PE can be transformed to KE within the system.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Potential energy (PE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

Furthermore, the total energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Total energy = P.E + K.E

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

In this scenario, a system has both potential energy (PE) and kinetic energy (KE).

According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

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