the researcher say hi for us the best pa the best of us are going out to eat that I can get my money toward a little bit but the best of luck to be at work by then and we will see what the status

7 0

3 years ago

HELP ASAP PLEASE!!!

blagie [28]

Answer: A

Explanation:Earthquakes occur on faults - strike-slip earthquakes occur on strike-slip faults, normal earthquakes occur on normal faults, and thrust earthquakes occur on thrust or reverse faults. When an earthquake occurs on one of these faults, the rock on one side of the fault slips with respect to the other.

4 0

2 years ago

Read 2 more answers

A 5.5 kg box is pushed across the lunch table. The acceleration of the box is 4.2 m/s2. What was the net force applied to the bo

Bezzdna [24]

Use this numbers that you are given and plug them into the formule for net force

Net force= m*a (mass times acceleration)

Therefore the net force is equal to (5.5 kg)*(4.2 m/s^2)= 23.1 N

Answer: Net force= 23.1 N

5 0

3 years ago

Read 2 more answers

ball dropped from top of 50m high cliff stone thrown straight up from bottom with speed of 24. they collide. how far above base

Tresset [83]

Answer:

28.73 m from the base of the cliff collide happen

Explanation:

Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by

$x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)$

stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)

$y=24t-\frac{1}{2} gt^2.............(2)$

we know that total distance traveled by ball and stone is 50 m

$x+y=50 m$

adding equation 1 and 2, we get time t

$x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s$

substitute this time in equation 2, we can get the required distance where they collide

$y=24t-\frac{1}{2} gt^2\\y=24\times 2.083-\frac{1}{2}\times 9.8\times 2.083^2\\y=28.73 m$

28.73 m from the base of the cliff collide happen

7 0

2 years ago