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alina1380 [7]
3 years ago
6

If a car is moving backward and has negative acceleration, what can be said about the speed of the car?

Physics
2 answers:
Alona [7]3 years ago
7 0
C? I really don’t know -.-
KATRIN_1 [288]3 years ago
4 0
I think its A let me know its wrong or not
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А man runs 20 km west is hows. He then runs 20km 60° north or west 2.5 hours. what's the average velocity the Runnner?​
shepuryov [24]

Explanation:

\large{ \star \fcolorbox{magenta}{purple}{ \fcolorbox{magenta}{blue}{ \fcolorbox{magenta}{pink}{ \fcolorbox{magenta}{red}{ \sf{ añswër}}}}}} \star

<h2>8km/h</h2>
7 0
2 years ago
What is the equation for determining the value of a specific gas constant using the universal gas constants
maria [59]

Answer:

Physically, the gas constant is the constant of proportionality that relates the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. Thus, the value of the gas constant ultimately derives from historical decisions and accidents in the setting of the energy and temperature scales, plus similar historical setting of the value of the molar scale used for the counting of particles.

Explanation:

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2 years ago
An umbrella tends to move upward on a windy day because _____.
masha68 [24]
E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
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3 0
3 years ago
A jet aircraft is traveling at 262 m/s in hor-
NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

   = 4862635.79 + 3610.32

   = $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$.

3 0
3 years ago
How much work is required to lift a 2 kilogram mass to a height of 10 meters
Semmy [17]

Given mass = 2kg, height = 10m,g = 9.8.


We know that Work done W = FD


= > W = (mg)(D)


= > W = (2 * 9.8)(10)


= > W = 196 Joules.


Hope this helps!

4 0
3 years ago
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