Answer:
V(t1-t0)
Explanation:
Moving 'uniformly' means constant velocity (speed). the formula for constant speed motion is
=( change in position/ change in time)
where,
V is speed
given in the statement :
change in time = t = t1-t0
let the constant speed be ' V '
disance = X = X1-X0
applying the above mentioned formula: V = 
V = X/t
X = Vt
the distance X1-X0 = Vt =V(t1-t0)
The highest elevation reached by the ball in its trajectory is 16.4 m.
To find the answer, we need to know about the maximum height reached in a projectile.
What's the mathematical expression of the maximum height reached in a projectile motion?
- The maximum height= U²× sin²(θ)/g
- U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity
What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?
- Here, U = 30.0 m/s and θ= 25°
- Maximum height= 30²× sin²(25)/9.8
= 16.4m
Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.
Learn more about the projectile motion here:
brainly.com/question/24216590
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Let's check the relationship


So
- Raindrops will fall faster . .
- Also walking on ground would become more difficult as g increases.
Option C is wrong by now .Let's check D once

- So time period of simple pendulum would decrease.
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:

where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
The answer would be B: The ground exerts an equal force on the golf ball.