1. How far can your little brother get if he can travel at 2.5 m/s and in 5 seconds

A track consists spring launcher on one end. A spring which is compressed 0.5 m has a

patriot [66]

(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.

(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.

<h3>Speed of the block when pushed by the spring</h3>

The speed of the block when pushed by the spring is calculated as follows;

K.E = Ux

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v² = (25 x 0.5²)/0.05

v² = 125

v = 11.18 m/s

<h3>Final velocity of the two balls after the collision</h3>

The velocity of the two balls after the collision is calculated as follows;

Pi = Pf

where;

Pi is initial momentum
Pf is final momentum

m1u1 + m2u2 = v(m1 + m2)

0.05(11.18) + 0.05(0) = v(0.05 + 0.05)

0.559 = 0.1v

v = 5.59 m/s

Learn more about velocity here: brainly.com/question/4931057

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4 0

2 years ago

Can you hellp me please

mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0

3 years ago

Read 2 more answers

A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts

galina1969 [7]

Answer:

$\mu_s = 0.62$

$\mu_k = 0.415$

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

$x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2$

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

$F = ma\\F = 1.7m$

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

$F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415$

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

$mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62$

5 0

3 years ago

A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object

Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

8 0

3 years ago

The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth),

SVETLANKA909090 [29]

Answer:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

$V_e=\sqrt{\frac{2GM}{R}}$

Escape velocity at the surface of earth

$11=\sqrt{\frac{2GM}{R}}$ --------------------1

If Escape velocity is three times and the radius is also the three times

$V_e_2=\sqrt{\frac{2G(3M)}{3R}}$

$V_e_2=\sqrt{\frac{2GM}{R}}=V_e$

i.e. $V_e_2=11 km/s$

5 0

3 years ago