I'm guessing they rounded because it is technically 2.998 X 108 miles/hour
so I would go with true!
Answer:
the second one i guess????
Explanation:
Answer:
D) The ball exerts a force on the wall and the wall exerts a force back.
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"
In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:
The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.
Answer:
The atmospheric pressure is 
.
Explanation:
Given that,
Atmospheric pressure 
drop height h'= 27.1 mm
Density of mercury 
We need to calculate the height
Using formula of pressure
Put the value into the formula
We need to calculate the new height
We need to calculate the atmospheric pressure
Using formula of atmospheric pressure
Put the value into the formula
Hence, The atmospheric pressure is .
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.
The x location therefore is 290*cos(53)*35 = 6108.4m
The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.
This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m
So your (x,y) coordinates equals (6108.4, 2097.5)
