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sp2606 [1]
3 years ago
12

If a 5kg ball is rolling to the left and has a momentum of -25kg*m/s what is the velocity (not just speed) of the ball

Physics
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

The velocity of the ball is 5 m/s to the left.

Explanation:

-25kg*m/s divided by 5 kg gives

speed of -5 m/s

The velocity of the ball is 5 m/s to the left.

You might be interested in
a 250g moves eastward along a straight path at a constant velocity of 12 m per second calculate the momentum of the ball ​
lara31 [8.8K]

Explanation:

If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .

6 0
3 years ago
Which of the following would decrease the magnetic field around a wire?
shusha [124]
The correct answer is
<span>A. decreasing the current

In fact, the magnetic field produced by a current carrying wire is given by
</span>B(r) =  \frac{\mu_0 I}{2 \pi r}
<span>where
</span>\mu_0 is the vacuum permeability
<span>I is the current in the wire
r is the distance from the wire at which the field is calculated

We see from the formula that the intensity of the field, B, is directly proportional to the current I, so if the current decreases, the magnetic field strength B decreases as well.</span>
4 0
3 years ago
Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
tankabanditka [31]

Answer:

a) 24.33 m of distance.

b) 34.55° east of the south.

Explanation:

The question is incomplete. The whole exercise is the following:

<em>"Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 28.0 m in a direction 60.0° west of north. Jane walks 12.0 m in a direction 30.0° south of west. They then stop and turn to face each other. </em>

<em>(a) What is the distance between them? </em>

<em>(b) In what direction should Ricardo walk to go directly toward Jane?</em><em>"</em>

Now that we know what we need to do in this question, let's head for every part of the problem.

<u>a) Distance between Ricardo and Jane</u>

In this case, we need to analyze the given data:

Ricardo (which we will call R) is 28 m from the starting point at 60° west of north, and Jane (J) is 12 m at 30° south of west. So the distance between them, will be the point where they both stop and face each other. This point can be seen in the image attached (See picture).

Let's call the distance between them as "D", to get the distance of D, according to the picture will be:

D = J - R   (1)

However, as they are facing in different angles and directions, we cannot do the difference of their values distance just like that. In order to do that, we need to calculate the components in the "x" and "y" axis of each vector. In that way, we can get the components of x and y of the Distance D, and then, the whole distance between them will be:

D = √Dx² + Dy²     (2)

So, let's get the components of x and y of R and J.

For Ricardo (R):

Rx = R sin60° = 28 sin60° = -24.25 m

Ry = R cos60° = 28 cos60° = 14 m

The sign "-" it's because R it's on the second quadrant, therefore in x, we'll have to add the negative.

For Jane (J):

Jx = J cos30° = 12 cos30° = -10.39 m

Jy = J sin30° = 12 sin30° = -6 m

Again, the negative is added because J is on the third quadrant.

Now that we have the components, let's calculate vector D using expression (1):

Dx = -10.39 - (-24.25) = 13.86 m

Dy = -6 - 14 = -20 m

Now, using expression (2) we can finally know the distance between Jane And Ricardo:

D = √(-20)² + (13.86)²

<h2>D = 24.33 m</h2>

This is the distance between Jane and Ricardo.

b) Direction of Ricardo walking to Jane

In this case, we already have the components of x and y of the distance between them, so, to know the direction:

Tanα = Dy/Dx

α = tan⁻¹ (Dy/Dx)

Replacing the values we have:

α = tan⁻¹ (-20/13.86)

α = 55.45°

Which should south of east or:

β = 90 - 55.45

<h2>β = 34.55°</h2>

Ricardo should walk 34.55° east of south

Hope this helps

8 0
3 years ago
Most of the mass of the Milky Way exists in the form of:<br> a.) starsb.) gasc.) dustd.) dark matter
Sveta_85 [38]
<h2>Answer: dark matter</h2>

It is believed that the Milky Way has 90% dark matter and only 10% ordinary matter (known matter). Because, like gravity, <u>dark matter can not be observed directly</u>, however its existence is inferred through the movement of the stars and the cosmic dust within the galaxy.

It is important to note that dark matter composition is unknown and corresponds to 80% of the matter in the universe. It does not emit or interact with any type of electromagnetic radiation, but it interacts with the known matter through gravity.  

5 0
3 years ago
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