Answer:3
Explanation:
First ball is thrown with horizontal velocity while other ball is dropped from cliff such that both have zero vertical velocity. So both balls have to cover a distance equal to the height of cliff with same initial velocity.
time taken is given by 
where h=height of cliff
g=acceleration due to gravity
horizontal velocity to first ball will make the ball to travel more horizontal distance as compared to second ball.
Option 3 is correct
Hi,Find answers from Task 5
1.(X+4)+(X)+(X+4)+(X)=50cm
4x+8=50cm
4x=42
X=10.5cm
Length=10.5+4=14.5cm
Width=10.5cm
Area= length × width=(10.5/100) × (14.5/100) =0.0152m2
2. Volume of a sphere= 4/3 ×π×r³
4/3 ×π×r³=3.2×10^-6 m³
r³=3.2×10^-6 m³/1.33×π
r³=7.64134761e-7
r=0.00914m
Surface area of the blood drop= 4πr²
=4×3.142×0.00914×0.00914=0.00105m²
3.
Equation of an ideal gas = PV =n RT
Equation for pressure, = P= n RT/V
Equation for the volume of an ideal gas= V= n RT/P
If the volume of gas doubles ,V(new)= 2n RT/P
Equation for temperature of an ideal gas, T = PV/n R
If temperature of gas triples, T (new)= 3PV/n R
New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)
Pressure factor increase= P(new)/P(old) ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}
=3PV²/2n RT
Answer:
Explanation:
The correct answer is Metabolic alkalosis (D). A pH of 7.48 shows slight alkalinity, this normal concentration of Co2 in the blood ranges from 35 mmHg (millimetre Mercury) to 45 mmHg and the normal HCo3 ( Hydrogen trioxo carbonate ion) concentration ranges from 22mEq/L to 26mEq/L.
Therefor the patients pH level is high the Co2 level is normal and the HCo3 level is high. Hence, Metabolic alkalosis
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
Answer:
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