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pav-90 [236]
3 years ago
6

4. A 180 kg sumo wrestler on roller skates, rolling at 5 m/s to the left, collides elastically with

Physics
1 answer:
Goryan [66]3 years ago
4 0

Answer:

m1u1+m2u2=m1v1+m2v2

Explanation:

m1=180

m2=215

u1=5

u2=3.5

v1=2

v2=?

use the above formula

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Fraunhofer single slit explanation
12345 [234]

Answer:

This is an attempt to more clearly visualize the nature of single slit diffraction. The phenomenon of diffraction involves the spreading out of waves past openings which are on the order of the wavelength of the wave.

Explanation:

4 0
3 years ago
1. Stops: Using the information learned in this course, explain three things you will not do when driving.
vazorg [7]
Don't text while driving
don't get your eyes off the road
don't get distracted
4 0
3 years ago
Find the average speed of a car that travels 45km in 2hours 15minutes​
Helga [31]

Answer:

20 km/h

Explanation:

45 km ÷ 2.25 hours (15 mins is 0.25 hours)

= 20

20 km/h

3 0
2 years ago
An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package
jeka94

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

8 0
3 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

 

4 0
3 years ago
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