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Shkiper50 [21]
4 years ago
15

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

way that would allow for a safe takeoff?

Physics
1 answer:
pickupchik [31]4 years ago
3 0
Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

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A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted
Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

F_c = F_g - F_N

where;

Fc is the centripetal force

F_g is downward force due to weight of the driver

F_N is upward or normal force on the drive

F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0
3 years ago
Graph are pictorical representations of
inn [45]

Answer:

Graphs are pictorial representations of relationships.

6 0
3 years ago
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POINTS WILL BE GIVEN!! VVV SIMPLE IF YOURE A GENIUS
insens350 [35]

Answer:

9

Explanation:

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6 0
3 years ago
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"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat
vagabundo [1.1K]

Answer:

(a) g = 8.82158145m/s^2.

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

g = \frac{M_{(earth)} }{r^2} G , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145m/s^2.

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

a_{centripetal}=\frac{V^2}{r} =g here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c)  S = vT,  here T is time period or time required to complete one full revolution.

S =  earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0
3 years ago
Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster
m_a_m_a [10]
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
4 0
3 years ago
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