CAN ANYONE PLEASE HELP ME WITH THIS ONE ITS URGENTLY NEEDED

Physics

1 answer:

denis-greek [22]3 years ago

4 0

Answer:

Explanation:

conservation of momentum

initial momentum is zero

Lets say that m₁ moves along the x axis

and m₂ moves along the y axis

in the y direction

0 = m₁(0) + m₂(30) + m₃(vy)

0 = m₁(0) + m₂(30) + 3m₂(vy)

-3m₂(vy) = m₂(30)

vy = -10 m/s

in the x direction

0 = m₁(30) + m₂(0) + m₃(vx)

0 = 2m₂(30) + m₂(0) + 3m₂(vx)

-3m₂(vx) = 2m₂(30)

vx = -20 m/s

v = √(-10² + -20²) = 10√5 m/s ≈ 22.36 m/s

θ = arctan(-10/-20) = 206.56505... ≈ 206.6° CCW from the + x axis

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For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex

rjkz [21]

Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches

1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches

4 0

3 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$