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qaws [65]
3 years ago
12

What affect results when there is an impact between earth and is astroids check all that apply.

Engineering
1 answer:
Vadim26 [7]3 years ago
6 0

Answer:

Massive destruction

Explanation:

If the asteroid collides with the ground, a massive volume of dust will be blasted into the environment. If it collides with water, the amount of water vapour in the atmosphere will rise. This would result in more rain, which would cause earthquakes and mudslides.

As the asteroid collided with the Earth, massive volumes of dust were ejected into to the atmosphere. The sun's light were stopped from entering the Earth's surface, which is terrible news for plants.

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Please help is due tonight
Kipish [7]

Answer:

tHE answer is b

Explanation:

7 0
3 years ago
Read 2 more answers
At a retirement party, a coworker described terry as dedicated
denis23 [38]

Answer:

At a retirement party, a coworker described Terry as dedicated, hardworking, and dependable. He also said that Terry was a great leader, knew the computer system, and kept the company's finances in order

8 0
3 years ago
Match the test to the property it measures.
Vinvika [58]

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

7 0
3 years ago
Write a program that removes all spaces from the given input. You may assume that the input string will not exceed 50 characters
GrogVix [38]

Answer:

Program that removes all spaces from the given input

Explanation:

// An efficient Java program to remove all spaces  

// from a string  

class GFG  

{  

 

// Function to remove all spaces  

// from a given string  

static int removeSpaces(char []str)  

{  

   // To keep track of non-space character count  

   int count = 0;  

 

   // Traverse the given string.  

   // If current character  

   // is not space, then place  

   // it at index 'count++'  

   for (int i = 0; i<str.length; i++)  

       if (str[i] != ' ')  

           str[count++] = str[i]; // here count is  

                                   // incremented  

         

   return count;  

}  

 

// Driver code  

public static void main(String[] args)  

{  

   char str[] = "g eeks for ge eeks ".toCharArray();  

   int i = removeSpaces(str);  

   System.out.println(String.valueOf(str).subSequence(0, i));  

}  

}  

5 0
4 years ago
Read 2 more answers
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 =  370.758 + 16^2

d^2 =   626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

5 0
3 years ago
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