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Goryan [66]
2 years ago
15

a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio

nary. As a result of the collision, the 2.5 g ball was given a velocity of 1.0 m/s along the same line. Calculate the speed and irection of the 0.80 kg ball immediately after the collision.​
Physics
1 answer:
Likurg_2 [28]2 years ago
5 0

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

  • mass of the ball, m₁ = 0.8 kg
  • speed of the ball, u₁ = 2.5 m/s
  • mass of the object at rest, m₂ = 2.5 kg
  • final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁  +  2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

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A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t
labwork [276]

Answer:

<h3>(A) The width x = 0.24 \times 10^{-3} m</h3><h3>(B) The new width is 1.32 \times 10^{-3} m</h3>

Explanation:

Given :

Focal length f =   135 \times 10^{-3}  m

Maximum aperture D = \frac{f}{4}

Wavelength \lambda = 550 \times 10^{-9} m

(A)

From rayleigh criterion,

  \theta = \frac{1.22 \lambda }{D}

  \theta =\frac{ 1.22 \times 550 \times 10^{-9}  }{33.75 \times 10^{-3} }

  \theta = 1.98 \times 10^{-5} rad

From angle formula,

  x = R\theta

Where R = 12 m ( given in example )

x = 12 \times 1.98 \times 10^{-5} m

x = 23.76 \times 10^{-5}

x = 0.24 \times 10^{-3} m

(B)

We know that \theta is proportional to the x and inversely proportional to the D

so we write the new width, here x is 5.5 times larger than above case

   x = 0.24 \times  10^{-3}  \times \frac{22}{4}

   x = 1.32 \times 10^{-3} m

6 0
2 years ago
Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small
ahrayia [7]
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by 

<span>C = 2L = 2*pi*R ---> R = L/pi </span>

<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>

<span>we can define a small charge dq as </span>

<span>dq = l*ds = l*R*da </span>

<span>So the electric field can be written as: </span>

<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>

<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
8 0
3 years ago
A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le
dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

8 0
2 years ago
Box A is pulled 2 meters when a 3-newton force is applied to it.
murzikaleks [220]

Answer: 0

Explanation: 1-1

3 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
2 years ago
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