Answer:
0.35 kg
Explanation:
8 cm = 0.08 m
For the block to stay balance, the buoyancy force must be the same as gravity that pulls it down.
Let mass of the block be M, then the gravity would be Mg
Let water density be 
, the buoyancy force would be the weight of water that is displaced by the submerged block.
For example, when there is no coin, block is 
submerged. The weight of water displaced must be
Which is also the weight of block, of Mg
Therefore M = 31.2A. (1)
As coins are stacked on top of block, h increase, so as weight of water displaced and total weight of block and coins. Now let m be the total weight of coins. The gravity of block and weight must be (M+m)g. And the weight of water displaced is:
Since the linear plot of h vs m has a slope of 0.089 m/kg, we can interpret it as
So from the eq. (1) we can solve for M = 31.2A = 0.35 kg
Answer:
1 V / div
Explanation:
Solution:
- The vertical scale has eight divisions.
- If each division is set to equal 1 volt, the display will show 0 to 8 volts.
- This is okay in a 0 to 5 volt variable sensor such as a throttle position (TP) sensor.
- The volts per division (V/div) should be set so that the entire anticipated waveform can be viewed.
The answer is 0
potential energy= mass• gravity (9.8m/s)• height.
Your mass 3kg. I believe your need to convert this to grams (3000g) and then multiply by 9.8. Since your height is 0 because the ball is on the ground then your potential energy is 0
Answer:
(1) 14.12 m/s
Explanation:
Given:
= initial speed of the ball = 16 m/s
= angle of the initial speed with the horizontal axis = 
= initial height of the ball from where Julie throws the ball = 1.5 m
= final position of the ball where Sarah catches the ball = 1.5 m
Let us assume the following:
= horizontal component of the initial speed
= vertical component of the initial speed
= horizontal acceleration of the ball
= vertical acceleration of the ball
The given problem is projectile motion. When the ball is thrown from the air with a speed of 16 m/s at an angle 28 degree with the horizontal axis. When the ball is in the air, it experiences an only gravitational force in the downward direction if we ignore air resistance on the ball.
This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.
Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.
Let us find out the initial velocity horizontal component of the velocity of the ball. which is given by:
As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.
Hence, the horizontal component of the ball's velocity when the ball is caught by Sarah is 14.12 m/s.
