It can either get brighter or even larger depends on the options given also
Answer:
16m
Explanation:
5m pluse 5m equals 10m 3m pluse 3m equals 6m 10m pluse 6m equals 16m thats ju ansar
Answer:14 m
Explanation:
Given
Vertical jump make by the dolphin is given by
Suppose the dolphin jump with an initial velocity of
so u is given by
If dolphin launches at an angle then maximum horizontal range is given by
assuming the of Dolphin to be Projectile so range is given by
substitute the value of
Range will be maximum for
thus
Answer:
and direction
Explanation:
because when we definite force force is the pull or push of an object and when a force is exerted on the object it works 3 tasks 1, it multiply the force , 2, it multiply the speed, 3, it changes the directions so when we see a vector quantity is a quantity that have both magnitude and direction so when it multiply it's force and speed the object will have magnitude and when it multiply it's direction the object will have a direction. thank you for reading this explanation.☺☺☺☺
Answer:
9.39 m/s
Explanation:
Using the y-direction, we can solve for the time t it takes for the cart to reach the ground.
Assume the up direction is positive and the down direction is negative.
- v₀ = 0 m/s
- a = -9.8 m/s²
- Δy = -50 m
- t = ?
Find the constant acceleration equation that contains these four variables.
Substitute known values into this equation.
Multiply and simplify.
Divide both sides of the equation by -4.9.
Square root both sides of the equation.
Now we can use this time t and solve for v₀ in the x-direction. Time is most often our link between vertical and horizontal components of projectile motion.
List out known variables in the x-direction.
- v₀ = ?
- t = 3.194382825 s
- a = 0 m/s²
- Δx = 30 m
Find the constant acceleration equation that contains these four variables.
Substitute known values into the equation.
- 30 = (v₀ · 3.194382825) + 1/2(0)(3.194382825)²
Multiply and simplify.
Divide both sides of the equation by 3.194382825.
The cart was rolling at a velocity of 9.39 m/s (initial velocity) when it left the ledge.