31.3m/s
Explanation:
Given parameters:
Mass of rock = 40kg
Height of cliff = 50m
Unknown:
Speed of rock when it hits ground = ?
Solution:
We are going to use the appropriate motion equation to solve this problem
The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.
Using;
V² = U² + 2gH
V = unknown velocity
U = initial velocity = O
g = acceleration due to gravity = 9.8m/s²
H = height of fall
since the initial velocity of the bodyg is 0
V² = 2gH
V= √2gH = √2 x 9.8 x 50 = 31.3m/s
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Answer:
The correct option is;
C. 1,715 m
Explanation:
We are given the information from the group of teen at the City edge
Time of arrival of explosion sound = 5 s after sighting
Time of sighting explosion = 5 s before hearing the boom
Speed of sound in air ≈ 343 m/s
Speed of light = 299,792 km/s
Therefore, distance covered by sound in 5 seconds is given by the following equation;


Hence Distance = 343 m/s × 5 s = 1715 m
To check, we compare the time it would take for the light to cover 1715 m
That is
which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.
Therefore, the distance of the students from the factory is approximately 1,715 m
Incomplete question as the unit of volume is not written correctly.So the complete question is here:
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³?
Answer:

Explanation:
Given data
Mass m=240g
Volume V=89.0 cm³
To find
Density d
Solution
If rock displaces 89.0 cm³ of water means volume of rock is also 89cm³
So

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