All categories

3 years ago

URGENT What does Newton's third law say about why momentum is conserved in collisions?

2 answers:

7 0

The correct answer is :
2. Equal forces act in equal times, so the change in momentum for both objects must be equal.

I just took the Test and it was correct

valentinak56 [21]3 years ago

4 0

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction

You might be interested in

Planets A and B have the same size, mass, and direction of travel, but planet A is traveling through space at half the speed of

Ganezh [65]

Answer:

B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.

Explanation:

Given that Planets A and B have the same size, mass.

Let the masses of the planets A and B are $m_A$ and $m_B$ respectively.

As masses are equal, so $m_A=m_B\cdots(i)$ .

Similarly, let the radii of the planets A and B are $r_A$ and $r_B$ respectively.

As radii are equal, so $r_A=r_B\cdots(ii)$ .

Let my mass is m.

As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.

So, my weight on planet A, $w_A= \frac {Gm_Am}{r_A^2}$

my weight of planet B, $w_B=\frac {Gm_Bm}{r_B^2}$

By using equations (i) and (ii),

$w_B=\frac {Gm_Am}{r_A^2}=w_A$ .

So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.

Hence, option (B) is correct.

4 0

3 years ago

A tank whose bottom is a mirror is filled with water to a depth of 20.1 cm . A small fish floats motionless 6.20 cm under the su

Studentka2010 [4]

Answer:

A) 4.66 cm

B) 25.56 cm

Explanation:

A) real depth of fish = 6.2 cm

apparent depth =d

Now $\frac{6.2}{d}$ = 1.33

$d = \frac{6.2}{1.33}$

d = 4.66 cm

B) Image of fish will be formed 13.9 cm behind the mirror so its position from

surface of water will be

20.1 + 13.9 = 34 cm . This will be real depth of image

If D be its apparent depth then

\frac{34}{D}=1.33

D =\frac{34}{1.33}

D = 25.56 cm

8 0

4 years ago

A student is skateboarding down a ramp that is 6.0 m long and inclined at 180 with respect to the horizontal. The initial speed

dlinn [17]

Answer:

The speed at the bottom of the ramp is 2.6m/s

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

Notice that it is necessary to found the acceleration that can be done by means of Newton's second law:

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

Force in the x axis:

$F_{x} = W_{x}$

The component of the weight in the x axis can be gotten by means of trigonometric:

$\frac{OC}{H} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$F_{x} = mgsen \theta$ (3)

Forces in the y axis:

$F_{y} = N - W_{y}$ (4)

The component of the weight in the y axis can be gotten by means of trigonometric:

$\frac{AC}{H} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

$N = mgcos \theta$

Therefore, equation 4 can be rewritten as:

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$ (5)

Then, replacing equation 3 and equation 5 in equation 2 it is gotten:

$mgsen \theta + 0 = ma$

$mgsen \theta = ma$ (6)

However, a can be isolated from equation 6

$a = \frac{mgsen \theta}{m}$

$a = gsen \theta$ (7)

Finally, equation 7 can be replaced in equation 2:

$v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}$

$v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}$

$v_{f} = 2.6m/s$

Hence, the speed at the bottom of the ramp is 2.6m/s

7 0

3 years ago

When a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b

Likurg_2 [28]

This is caused by A. resonance.
hope this helps!

5 0

4 years ago

Read 2 more answers

I just... I don't know man I suck at physics :(

lys-0071 [83]

Answer:

Explanation:

The velocity is 340, but the distance is not. Here's how you do this:

You need to find out the velocity of the that sound in that temp which has a formula of

v = 331.5 + .606T where T is the temp in Celcius. Filling in and using the correct number of sig dig:

v = 331.5 + .606(20) and

v = 331.5 + 10 and

v = 340 m/s. Now we need to use that in d = vt:

d = 340(2.00) so

d = 680 m

3 0

3 years ago

Read 2 more answers