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4 days ago

Which statements correctly describe forces? Check all that apply.

Physics

2 answers:

Yuliya22 [1.1K]4 days 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 [1K]4 days ago

4 0

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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The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

ValentinkaMS [1144]

Answer:

The force magnitude is $F_{net}= 1.837 *10^4N$

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, $\theta = 20^o$

The angular rate of increase is $w = 2 \ ^o/s$

Converting to revolutions per second gives us $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The rope length is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , Tension T of the rope is 18 kN.

The weight of the para-sailor is $M_p = 75kg$

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

The derivative of displacement results in velocity.

Hence,

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

signifies the velocity, and further differentiation yields acceleration.

Therefore,

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

$w = \frac{\Delta \theta}{\Delta t}$

where $\theta$ represents the angular displacement.

Next, evaluating the interval from $20^o \ to \ 30^o$ gives us

$2 = \frac{30 -20 }{t -0}$

making t the focal point.

$t = \frac{10}{2}$

$= 5s$

At this time, the displacement measures

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity computes to

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

Whereas linear acceleration calculates as

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally, radial acceleration is given by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Simultaneously, angular acceleration can be represented as

$\alpha_t = r \theta'' + 2 r' \theta '$

Then $\theta '' = \frac{d (0.0349)}{dt} = 0$

Thus,

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The resultant acceleration is mathematically denoted as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the acceleration's direction is mathematically expressed as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The y-axis force acting on the para-sailor is mathematically shown as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The x-axis force acting on the para-sailor is represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The overall force is calculated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The directional force is evaluated as

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

1 day ago

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u

Yuliya22 [1153]

Answer:

The total energy can be expressed as $T = 169.02 \ J$

Explanation:

The problem states that

The Poynting vector, which measures energy flux, equals $k = 0.939 \ W/m^2$

The rectangle's length is represented by $l = 1.5 \ m$

The width of the rectangle is $w = 2.0 \ m$

The duration considered is $t = 1 \ minute = 60 \ s$

Mathematically, the overall electromagnetic energy incident on the area is given by

$T = k * A * t$

where A denotes the area of the rectangle, calculated as

$A= l * w$

By plugging in the respective values

$A= 2 * 1.5$

$A= 3 \ m^2$

Again substituting values

$T = 0.939 * 3 * 60$

$T = 169.02 \ J$

5 0

6 days ago

A compressed spring has 16.2J of elastic potential energy when it is compressed 0.30m . What is the spring constant of the sprin

ValentinkaMS [1144]

I hope this provides the assistance you need.

3 0

10 days ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [913]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

13 days ago

Which statements about inertia and centripetal force are correct? Check all that apply. Inertia is always present. Inertia cause

kicyunya [1011]

Inertia is universally present. It's important to note that inertia doesn't serve as the force keeping objects in circular paths; that role belongs to centripetal force, which is not always present. Centripetal force actively pulls objects towards the center of a circle. Both inertia and centripetal force contribute to the phenomenon of circular motion. Thank you, and enjoy your day;)

6 0

5 days ago

Read 2 more answers