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Misha Larkins [42]
3 years ago
8

30 POINTS!

Physics
2 answers:
photoshop1234 [79]3 years ago
4 0

"D is correct answer." Velocity is the displacement of an object during a specific unit of time. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"

amm18123 years ago
3 0

Answer:

D

Explanation:

VELOCITY!

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During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi
vladimir2022 [97]

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

7 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
HELP ASAP ILL GIVE BRAINLIST
OlgaM077 [116]

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

4 0
2 years ago
Read 2 more answers
Only negative charges move<br><br> True<br><br> False
Flura [38]
The answer is False, pls. mark me the brainliest if I’m right. THX
7 0
3 years ago
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A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe
natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa =  709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

\frac{p}{T}  = constant

\frac{p_{1} }{T_{1} }  = \frac{p_{2} }{T_{2} }

p_{2}  = \frac{T_{2} }{T_{1} } *p_{1}  

p_{2}  = \frac{278}{300}  * 709.275  = 657.2615 Kpa = 6.486 atm

8 0
2 years ago
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