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3 years ago

In a collision between two unequal masses, which mass receives a greater magnitude impulse?

1 answer:

8 0

<span>The answer to this question depends upon Newton's third law of motion. For every action, there's an equal and opposite reaction. Because of this law, during the collision between two unequal masses, the impulse that each mass receives will be of equal magnitude and and opposite sign.</span>

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When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.

3 0

3 years ago

Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector $=7cos30^{\circ}=7\times 0.866=6.06$

y component of the vector $=7sin30^{\circ}=7\times 0.5=3.5$

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector $=7cos120^{\circ}=7\times -0.5=-3.5i$

y component of the vector $=7sin120^{\circ}=7\times 0.866=6.06j$

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0

4 years ago

A 190 g air-track glider is attached to a spring. The glider is pushed in 9.20 cm and released. A student with a stopwatch finds

Sholpan [36]

Answer:

$k = 12.136\,\frac{N}{m}$

Explanation:

The angular frequency of the system is:

$\omega = \sqrt{\frac{k}{m} }$

The frequency is:

$f = \frac{14\,osc}{11\,s}$

$f = 1.272\,hz$

The angular frequency is:

$\omega = 2\pi\cdot (1.272\,hz)$

$\omega = 7.992\,\frac{rad}{s}$

The spring constant is:

$k = \left(7.992\,\frac{rad}{s} \right)^{2}\cdot (0.190\,kg)$

$k = 12.136\,\frac{N}{m}$

6 0

3 years ago

Which scenario is an application of Newtons Second Law of Motion?

svlad2 [7]

D would be the answer because The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

4 0

3 years ago

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length

grin007 [14]

Answer:

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

The correct answer is

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5 = vb² +2g×vbt+1/2gt²

Explanation:

We have from the relation

v = u + gt, S = ut + 1/2 gt², v² = u² + 2gS

in this case S = hb, u = vb=Lwt+gt2. and v = vground

therefore v² = (Lwt+gt²)² + 2 × g × hb

= (Lwt+gt²)² + 2 × g × (Lwt+gt²)×t + 1/2 gt² = vb² +2g×vbt+1/2gt²

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5

7 0

3 years ago