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3 years ago

Anyone?????help with this questions!!!

Physics

2 answers:

Sergio039 [100]3 years ago

7 0

Hey there!

A) Drag (Drag Force) is a force that acts on the opposite side to the motion that an an object is traveling in. For example, the drag force on an airplane would be at the front because that’s opposite to the direction of motion.

B) To Increase the Thrust.

Vlada [557]3 years ago

4 0

Answer:

A) Drag is when a resistive force (in this case, air resistance) causes the moving vehicle to decrease its speed.

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A force of 13 N is used to slide a 0.8 kg book across a table. What is the magnitude of the book’s acceleration?

Mkey [24]

Explanation:

acceleration=F/m

a=13/0.8

a=16 25m/s²

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4 years ago

The molecular weight of oxygen gas (o2) is 32 while that of hydrogen gas (h2) is 2.

anzhelika [568]

use the formula: v^2=(3kT)/m

Where:

v is the velocity of a molecule

k is the Boltzmann constant (1.38064852e-23 J/K)

T is the temperature of the molecule in the air

m is the mass of the molecule

For an H2 molecule at 20.0°C (293 K):

v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)

v^2 = 3.65e+6 m^2/s^2

v = 1.91e+3 m/s

For an O2 molecule at same temp.:

v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)

v^2 = 2.28e+5 m^2/s^2

v = 478 m/s

Therefore, the ratio of H2:O2 velocities is:

1.91e+3 / 478 = 4.00

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3 years ago

Which statements accurately describe mechanical waves

AleksAgata [21]

Answer:

Explanation:

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3 years ago

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

⠀

$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

⠀

The horizontal range of projectile at x.

⠀

$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

⠀

$\large\textsf{\underline{Now substituting the required values}}$

⠀

$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

⠀

The vertical position of projectile at y.

⠀

$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

⠀

$\textsf{ \large {\underline{Now substituting the required values}} }$

⠀

$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$