What is the net force acting on a .15 kg hockey puck accelerating at a rate of 12 m/s2

Physics

2 answers:

maw [93]3 years ago

7 0

Answer:

The net force is 1.8N

Explanation:

Given that the formula for force is Force = mass×acceleration. So you have to substitute the values into the formula :

$force = mass \times acceleration$

Let mass = 0.15kg,

Let acceleration = 12m/s²,

$force = 0.15 \times 12$

$force = 1.8$

larisa [96]3 years ago

4 0

Answer:

1.8 Newtons

Explanation:

We want to figure out the net force acting on a 0.15 kg hockey puck that has an acceleration of 12 m/s². To solve this, we must apply Newton's Second Law of Motion which is denoted by the formula: f = ma.

Here, f is the force, m is the mass, and a is the acceleration. Plug our given values in to find f:

f = ma

f = (0.15 kg) * (12 m/s²) = 1.8 Newtons

Thus, the answer is 1.8 N.

~ an aesthetics lover

You might be interested in

A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an

kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

mass of the car, m = 1000 kg
speed of the car, v = 50 km/h = 13.89 m/s
banking angle, θ = 20⁰

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

Divide the second equation by the first;

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$