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Pie
3 months ago
10

Which statements correctly describe forces? Check all that apply.

Physics
2 answers:
Yuliya22 [3.3K]3 months ago
8 0

1. Forces can change the motion of an object.

TRUE

According to Newton’s first law, when an unbalanced force acts on an object, it will change the object's motion.

2. Forces are vectors.  

TRUE

As forces require both a direction and a magnitude, they are classified as vector quantities.

3. Forces are measured in joules.

FALSE

Forces are quantified in Newtons in the SI system, whereas Joules measure energy.

4. Forces are quantified in newtons.

TRUE

Forces are expressed in Newtons, honoring Sir Isaac Newton.

5. Forces can take on positive or negative values to show direction.

TRUE

As a vector quantity, forces can represent positive or negative to indicate direction.

Sav [3.1K]3 months ago
4 0
  1. Forces can alter an object's motion

Indeed, this is unquestionably TRUE. At this moment, I am applying force to my keyboard, affecting the state of motion of the keys. Therefore, it is factual that forces can change an object's state of motion.

2. Forces are vectors

Yes, this statement is TRUE. Forces are considered vector quantities as they possess both magnitude and direction. Describing a force requires knowing its size and direction.

3. Forces are quantified in Joules

It depends, are we discussing physical force or electrical force? In this context, we are focusing on physical forces, given that the query does not mention electrical ones. Forces are typically measured in Newtons (N) (kg*m/s^2). Hence, this assertion is FALSE.

4. Forces are quantified in Newtons

This concept has appeared before! This assertion is TRUE. (assuming the question is not regarding electrical forces).

5. Forces can have positive or negative values to denote direction

This is TRUE, and I have failed numerous tests because of this. Keep in mind, the negative sign indicates direction rather than a negative quantity! Be sure to read test instructions to see if direction needs to be included.


Physics is enjoyable!



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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
3 months ago
A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Sav [3153]
The total duration from the stone's drop to the echo being heard is 8.9 seconds. The sound takes 0.9 seconds to reach the listener, meaning it takes the stone 8 seconds to descend to the well's bottom. Using gravitational acceleration, the well's depth calculates to be approximately 313.6 meters.
4 0
2 months ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [3204]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0
3 months ago
Read 2 more answers
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