B. Variable
C. Control
D. Hypothesis
c. What is the instantaneous velocity of the rock just before it hit the ground?
d.Find the rock's displacement for the 10 seconds.
c. -98 m/s
The motion of the rock is a uniformly accelerated motion (free fall) with constant acceleration (negative since it is downward). Therefore, we can find its velocity using the following suvat equation
where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity
t is the time
For the rock in the problem,
u = 0
So, its velocity at t = 10 s is
where the negative sign indicates that the velocity points downward.
d. -490 m
Since the motion is at constant acceleration, we can use another suvat equation:
where
s is the displacement
u is the initial velocity
g is the acceleration of gravity
t is the time
Substituting:
u = 0
t = 10 s
We find the rock's displacement:
where the negative sign means the displacement is downward.
B. One space probe has more air resistance than the other.
C. Only one space probe is exerting a gravitational force on the other.
D. One space probe is closer to Jupiter than the other
Answer:
The correct answer is D
Explanation:
B.)delta
C.)valley
D.)floodplain
Answer:
D
Explanation:
Answer:Static Electricity
Explanation:
Answer:
The horizontal component of this force is about 36 N.
Explanation:
Given that,
Bruce is is pulling on his lead with a force of 40 N
Angle below horizontal,
We need to find the horizontal component of this force. The horizontal component of any vector is given by :
or
So, the horizontal component of this force is about 36 N. Hence, this is the required solution.