How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
1 answer:
Answer:
21 m
Explanation:
The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using
where
d is the distance covered
v is the speed
t is the time
The frog in this problem has a speed of
v = 2.1 m/s
and therefore, after t = 10 s, the distance it covered is
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Answer:
Compression of the spring: 0.18 m (downward)
Explanation:
The forces acting on the block of wood are:
- The force of gravity, acting downward, of magnitude , where m = 5.32 kg is the mass of the block and
is the acceleration due to gravity
- The force exerted by the spring, downward, of magnitude , where
is the spring constant and
is the elongation of the spring
- The buoyant force, upward, of magnitude , where
is the water density and V the volume of the block
Since the block is in equilibrium, the net force is zero, so we can write
(1)
We have to find the volume of the block first. We have:
m = 5.32 kg (mass)
(wood density)
So, the volume is
So now we can re-arrange eq.(1) to find the elongation of the spring, x:
So, the spring is compressed by 0.18 m.
Answer:
The total work that the rope does to Mangnus is - 5780 Jules.
Explanation:
By definition, the work is defined as:
Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.
Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.
For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement: