Scientific question must be

Physics

1 answer:

Rama09 [41]3 years ago

8 0

Are there any choices with the question?

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Nzvhybprob it is really coll to do it

nika2105 [10]

Answer:

gddtufxhjdtigdtuh abi na

6 0

3 years ago

Find the sum of the vectors:11 km N ,11km E

vladimir1956 [14]

The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

=11√2 km

Now for the direction, one vector is due north and the other is due east.

So the resultant vector is due north east.

So the final vector is 11√2 km due North-East.

6 0

3 years ago

A 1420-kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring

zzz [600]

Answer:

1419.01436 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-12.4^2}{2\times 78}\\\Rightarrow a=-0.98564\ m/s^2$

The force on the car

$F=ma\\\Rightarrow F=1420\times -0.98564\\\Rightarrow F=-1419.01436\ N$

Magnitude of the horizontal net force that is required to bring the car to a halt is 1419.01436 N

3 0

3 years ago

Read 2 more answers

In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is

goldfiish [28.3K]

In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is 600 J.Option c is correct.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

The displacement value from the triangle is;

$\rm sin 30^0 = \frac{P}{H} \\\\ H=S =\frac{3 \ m}{sin 30^0} \\\\ S = 6\ m$