The answer is "Three".
At the point when individuals take studies for factual purposes or when the Statistics Agency comes around and gets everyone's data, data from one individual is one information point for them. Toward the finish of its exploration or overview, the organization will have accumulated numerous bits of data from numerous individuals. One piece of information equals with one data point.
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
Answer:
1. all of them
2. cork and wax
3. iron, lead, and aluminum
4. none of them
Explanation:
1.Which material will displace a volume of water? all of them
When an object is introduced into a container with a volume of water, a volume of liquid equal to the volume of the object is displaced
2.Which material will displace a volume of water less than its own volume?
cork and wax
because the density of the object is less than that of the displaced liquid
3.Which material will displace a volume of water equal to its own volume?
iron, lead, and aluminum
because Arquimedes's principle: any body plunged inside a fluid in this case water experiences an ascending force called push, equivalent to the weight of the fluid removed by the body
4.Which material will displace a volume of water greater than its own volume?
None of them
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
