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4 years ago

Which of the following is a noncontact force? A. electrical forces B. gravitational forces C. magnetic forces D. all of these

Physics

2 answers:

Tamiku [17]4 years ago

7 0

Answer:

D. all of these

Explanation:

Electric forces, gravitational forces and magnetic forces are non contact forces. These forces are also the fundamental forces. Non contact forces are the forces which applied on a body without any real touch. Generally these non contact forces are in equilibrium.

goblinko [34]4 years ago

6 0

D. all of these

hope this helps.

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Do you think most people follow the recommended guidelines for an adequate exercise routine?

blondinia [14]

Personally, I know that I don't however it all depends on the person's lifestyle, health, and resoucres, such as if they are able to complete the excersise or have enough money fore equpemnt. To be safe, I would say that the majority of people do follow the guidelines although they may manipulate it to fit their best interest.

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3 years ago

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

just olya [345]

Force is defined as the rate of change of momentum.
The initial amount of momentum is $mv$ because water stops when it hit the wall total change of momentum must be $\Delta p=mv$ .
Now let's calculate the force.
$F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v$
We need to find $\frac{dm}{dt}$ . This is the amount of water hiting the wall per second.
$\frac{dm}{dt}=\rho Av$
Our final formula would be:
$F=\rho Avv=\rho Av^2$
And now we can calculate the answer:
$F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N$

6 0

3 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

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2 years ago

The tactile system is actually made up of two distinct skin senses.

Ludmilka [50]

The tactile system is actually made up of two distinct skin senses; protective and the discriminative system. The function of the protective system is to alert us to harmful stimuli. The receptors of the system are the hairs of our skin, head and genitals. The discriminative system is to show us what we are touching, where on our bodies is touching us and what are touching us. The receptors for his system are the hands, fingers, soles of feet and mouth and tongue.

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3 years ago

Read 2 more answers

Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin

andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

$W=Fd$

Substitute the value in the above equation.

$\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}$

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.