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3 years ago

a jetliner goes from rest to a takeoff speed of 162 mi/h in 39.5 s. what is the magnitude of its average acceleration?

1 answer:

3 0

Acceleration = V/t

First set velocity to m/s

V = 72.42 m/s

Now we divide velocity by time

A = 72.42/39.5
A = 1.83 m/s^2

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An object initially at rest experiences an acceleration

lidiya [134]

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

Average velocity = $\frac{total distance }{time }$

Total time taken = 5s + 6s = 11s

The first distance covered = velocity x time = 1.4 x 5 = 7m

second distance covered = velocity x time = 1.4 x 6 = 8.4m

So;

Average velocity = $\frac{7 + 8.4}{11}$ = 1.4m/s

5 0

3 years ago

The boiling point of water is 91.30 °C on the

Taya2010 [7]

Answer:

196.34 °F

Explanation:

To convert from degrees celsius to degrees fahrenheit, use this equation:

(°C * 9/5) + 32 = °F

So, using this equation:

(91.30 * 9/5) + 32 = °F

196.34 + 32 = °F

°F = 196.34

Hope this helps!

3 0

3 years ago

A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250

Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

= 6 hours

Total distance traveled = 350 km + 250 km

= 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

8 0

3 years ago

In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:

vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

Explanation:

[H⁺] [OH⁻] = 10⁻¹⁴

(1×10⁻⁶) [OH⁻] = 10⁻¹⁴

[OH⁻] = 1×10⁻⁸

6 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago