Answer:
The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.
Explanation:
Hope this helps Plz mark brainliest
convert 100 Newton into dyne
Answer:10000000
Explanation:
A golf ball flies through the air after being struck with a golf club. Which of the following statements describes the force on the ball as momentum is transferred between the club and ball?
A. The ball does not experience any force.
B. The force experienced by the ball is greater than the force experienced by the club.
C. The force experienced by the ball equals the force experienced by the club.
D. The force experienced by the ball is weaker than the force experienced by the club.
Answer:
C) The force experienced by the ball equals the force experienced by the club.
Explanation:
When the golfer strikes the ball with his club, the club exerts a force on the ball. Due to Newton's Third Law Of Motion [Every Action has an equal and opposite reaction], the ball also exerts an equal force on the club. However,
As the mass of the club is usually greater than the mass of the ball, it accelerates slower; While the ball way faster, following the equation : F=ma
A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction between the refrigerator and the floor is 230 N. How much work has been performed by the mover on the refrigerator?
Given :
A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.
The force of friction between the refrigerator and the floor is 230 N.
To Find :
How much work has been performed by the mover on the refrigerator.
Solution :
Since, refrigerator is moving with constant velocity.
So, force applied by the mover is also 230 N ( equal to force of friction ).
Now, work done in order to move the refrigerator is :
[tex]W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m[/tex]
Hence, this is the required solution.
What is the acceleration of the the object during the first 4 seconds?
Answer:
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
Explanation:
What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .
Answer:
It’s a
Explanation:
Don’t actually put that i needed the points mb
Which objects cannot be observed in detail without a microscope?
Answer:
partecls
Explanation:
because they are to small to see with plain eyes
Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Answer:
The value is [tex]p = 0.7556 c[/tex]
Explanation:
From the question we are told that
The speed at which galaxy B moves away from galaxy A is [tex]v = 0.577c[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
The speed at which galaxy C moves away from galaxy B is [tex]u = 0.731 c[/tex]
Generally from the equation of relative speed we have that
[tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]
Here p is the velocity at which galaxy C recede from galaxy A so
[tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]
=> [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]
=> [tex]0.731c - 0.4218 p = p - 0.577c[/tex]
=> [tex]0.731c + 0.577c = p + 0.4218 p[/tex]
=> [tex]1.308 c = 1.731 p[/tex]
=> [tex]p = 0.7556 c[/tex]