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3 years ago

What is the final velocity of a 10 kg object that drops from rest and falls for 30 seconds?

1 answer:

8 0

The object is in free fall, so its velocity follows the law:
$v(t) = v_0 + gt$
where
$v_0$ is the initial velocity (zero)
g is the gravitational acceleration
t is the time

Since the initial velocity is zero, the velocity after t=30 s is
$v(30 s)=gt=(9.81 m/s^2)(30 s)=294.3 m/s$

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3 years ago

In which situation is no work considered to be done by a force?

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3 years ago

If the potential is given by v = xy - 3z-2, then the electric field has a y-component of?

kiruha [24]

If the potential is given by v = xy - 3z-2, then the electric field has a y-component of X

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Each location in space where a charge exists in any form can be considered to have an electric field attached to it. The electric force per unit charge is another name for an electric field. The electric field's equation is given as E = F / Q. Volts per meter (V/m) is the electric field's SI unit. Newton's per coulomb unit is the same as this one.

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2 years ago

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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3 years ago