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2 years ago

(50 POINTS) What types of transportation have been modified over the years to decrease

Physics

1 answer:

kow [346]2 years ago

8 0

Answer:

planes gliders helecopters and rockets

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A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

pav-90 [236]

F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs

4 0

2 years ago

Read 2 more answers

Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 $m/s^2$

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

$v = u + gt$

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

$49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs$

The arrow was in flight for 3.8 secs

6 0

3 years ago

What burns quicker, white or colored candles?

VashaNatasha [74]

White is a hotter color so it burns faster. :)

3 0

2 years ago

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

2 years ago

Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc

expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0

3 years ago