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2 years ago

A Plane crashes into the side of a mountain. If the crash took place over 9

Physics

1 answer:

Julli [10]2 years ago

6 0

Given parameters:

Duration of crash = 9min

Velocity before crash = 2000 yards per second

Unknown:

Acceleration of the plane = ?

Solution:

Acceleration is the change of velocity with time.

Acceleration = $\frac{V - U}{t}$

where v is the final velocity

u is the initial velocity

t is the time taken

We need to convert the time into seconds from min;

60s = 1 min

So, 9min = 60 x 9 = 540s

So input the parameters and solve;

Acceleration = $\frac{2000 - 0 }{540}$ = 3.7yards/s²

The acceleration of the plane is 3.7yards/s²

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Two circular loops of wire, each containing a single turn, have the same radius of 5.10 cm and a common center. The planes of th

Fofino [41]

Explanation:

Below is an attachment containing the solution.

8 0

2 years ago

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$

Therefore, the particle's charge-to-mass ratio is $958.1\times10^5C/kg$

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0

3 years ago

Observations of galaxies and clusters of galaxies indicate that about ____ percent of the matter in the universe is dark matter.

Setler79 [48]

All the stars, spheres and galaxies that can be perceived nowadays make up just 5 percent of the universe. The former 95 percent is prepared of stuff stargazers can't see, notice or even understand. These secretive substances are called dark energy and dark matter. Experts determine their presence grounded on their gravitational influence on what little bits of the universe can be perceived.

5 0

2 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

3 years ago

A certain quantity of a liquid has a volume of 10cm at 20°C. Calculate its volume at 50°C, if it’s cubic expansivity is 10^(-3)

Pavel [41]

Answer:

10.3 cm³

Explanation:

From the question given above, the following data were obtained:

Original volume (V₁) = 10 cm³

Initial temperature (θ₁) = 20 °C

Final temperature (θ₂) = 50 °C

Cubic expansivity (γ) = 10¯³ K¯¹

Final volume (V₂) =?

γ = V₂ – V₁ / V₁(θ₂ – θ₁)

10¯³ = V₂ – 10 / 10( 50 – 20)

10¯³ = V₂ – 10 / 10(30)

10¯³ = V₂ – 10 / 300

Cross multiply

10¯³ × 300 = V₂ – 10

0.3 = V₂ – 10

Collect like terms

0.3 + 10 = V₂

10.3 = V₂

V₂ = 10.3 cm³

Thus, the volume at 50 °C is 10.3 cm³

6 0

2 years ago