Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his:

a. acceleration in his direction of motion
b. acceleration opposite to his direction of motion.

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.

In Non-Impact printers, no mechanical moving component is used.

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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:

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:

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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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

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:

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:

8760 N

Explanation:

think this is the right answer :)

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

Answer:

me...:(

Explanation:

Answer:

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

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]

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

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

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: 55 kmph

Explanation: Divide 110 by 2

Answer:

Explanation:

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

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:

[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:

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:

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.

Answer: 10 J

Explanation:

Trust ;)