Answer:
a) 42.9 m/s
b) 41.6m
Explanation:
a)
solving the eq we get
v=42.9 m/s
b) <em>here,</em>
when put in y(t) it gives y=44.6 m so result is
h=44.6-3= 41.6m
The final velocity of the Maserati after accelerating at the rate of 85 m/s² for 5 seconds is 431m/s.
<h3>How to calculate the final velocity a moving object?</h3>
From the first equation of equation of motion, final velocity is the sum of the initial velocity and the product of acceleration and time.
It is expressed as;
v = u + at
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
Given the data in the question;
- Initial velocity u = 6m/s
- Acceleration a = 85m/s²
- Elapsed time t = 5s
- Final velocity v = ?
Plug the given values into the first equation of motion and solve for v.
v = u + at
v = 6m/s + ( 85m/s² × 5s )
v = 6m/s + 425ms/s²
v = 6m/s + 425m/s
v = 431m/s
The final velocity of the Maserati after accelerating at the rate of 85 m/s² for 5 seconds is 431m/s.
Learn more about the first equation of equation here: brainly.com/question/20381052
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Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, must be in the north direction.
Let is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector and head of the vector must lie on the north-south line.
Now, for this condition, from the triangle OAB
Hence, the kayaker must paddle in the direction of in the north of east direction.
The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.