Ask question

Ask question

All categories

3 years ago

6

A particle is being accelerated through space by a 10 N force. Suddenly the particle encounters a second of 10 N force in the op

posite direction. The particle with both forces acting?

1 answer:

Lyrx [107]3 years ago

7 0

Answer:

The particle will be rendered still

Explanation:

The particle will be rendered still because of the Newton’s 1st law of motion states that a body will continue to move on forever until contacted by an equal and opposite force. The same case has been produced here and it is justifying the law. If the Force has another direction and magnitude than 10 N then the body will continue to move on in the direction of the greater force.

You might be interested in

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0

3 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago

The surface temperature of our sun is about 5800 K, and the peak of its intensity curve is in the middle of the visible spectrum

amm1812

Explanation:

i expected to use Stefan's law of heat exchange but the value you gave aren't conclusive.

I should say that the temperature of the star should be close to that of the sun because of the similarity in the intensity curves

3 0

2 years ago