Which of the following is a scalar? centripetal force weight speed acceleration

Physics

2 answers:

andreyandreev [35.5K]2 years ago

7 0

Answer:

Speed

Explanation:

- A scalar quantity is defined as a quantity which has a magnitude only (no direction)

- A vector quantity is defined as a quantity which has both a magnitude and a direction

Let's analyze each option:

- Centripetal force: force is a vector, because it has a magnitude and also a direction

- Weight: weight is a force, and force are vector quantities, so weight is a vector as well

- Speed: speed is a scalar quantity, because it has only a magnitude, no direction

- Acceleration: acceleration is a vector quantity, because it has a magnitude, but also a direction (because it represents the rate of charge in velocity, and since velocity is a vector, acceleration must be a vector as well)

So, the correct choice is

Speed

choli [55]2 years ago

6 0

Scalar is a quantity such as mass or length that is completely determined by its own magnitude and has no direction. Therefore, your answer is Speed.

You might be interested in

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$