Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
Answer:
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
Explanation:
The speeds of the particles are very large and comparatively the average free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.
Is proportional to the CHANGE.
hope this helps
First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton
Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons
Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)
Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg
Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2
Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters
Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec
Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec
Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:
Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:
- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
