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3 years ago

12

Suppose an object is moving 2.1 m/s north on a river, but the river is flowing to the east at a velocity of 1.2 m/s. What is the

magnitude of the resultant velocity?

1 answer:

rewona [7]3 years ago

8 0

Answer:

2.4 m/s

Explanation:

Given:

Velocity of the object moving north = 2.1 m/s

Velocity of the river moving eastward = 1.2 m/s

The resultant velocity is the vector sum of the velocities of object and river.

Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.

The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.

The magnitude of the resultant velocity (R) is given as:

$R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s$

Therefore, the resultant velocity has a magnitude of 2.4 m/s.

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2 years ago

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

2 years ago

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

Radda [10]

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

<h3>0.59 m/s²</h3>

Hope this helps you

7 0

3 years ago