A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?

Physics

2 answers:

damaskus [11]1 year ago

6 0

P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s

Tom [10]1 year ago

3 0

Magnitude:

√(3² + 4²) = √(9 + 16) = √25 = 5

Momentum = (mass) x (velocity) = (20kg x 5m/s) = 100 kg-m/s

Direction:

3i - 4j is in the direction tan⁻¹(-4/3) = 53.1 south of east.

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A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the

zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂= L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0

2 years ago

Read 2 more answers

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

Brums [2.3K]

We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
$S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm$