All categories

3 years ago

Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .

Physics

1 answer:

Radda [10]3 years ago

7 0

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{160000}{270000} = \frac{16}{27} \\ = 0.592592...$

We have the final answer as

Hope this helps you

You might be interested in

show how three identical 6 resistors must be connected tho have the following effective resistance values 9 and 4 ohms

SpyIntel [72]

Answer:

connect two 9 ohms resistance in series now it becomes 18 ohm

5 0

2 years ago

PLEASE HELPPPPP!!Calculate the net force acting on obiect A. In your answer, be sure to include the number, unit, and

asambeis [7]

Answer:

4 N to the right

Explanation:

For the object in this problem, we have 2 forces acting on it:

- A force of 2 N, to the right

The two forces are in line: this means that the two vectors can be simply added by calculating their algebraic sum.

If we take right as positive direction, the two forces can be written as

F1 = +2 N

F2 = +2 N

So, the net force will be

$F=F_1+F_2=+2+2=+4 N$

And the positive sign means the direction is to the right.

6 0

3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$