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Wewaii [24]
3 years ago
8

A boat is traveling with a velocity of 18 meters/second relative to water and the river is flowing at a velocity of 2.5 meters/s

econd relative to the shore. What will be the velocity of the boat if the boat and the water are in opposite directions?
Physics
2 answers:
ddd [48]3 years ago
8 0
Vt = Vboat - Vriver 
Vt = 18 - 2.5 = 15.5 m/s 

If the boat's direction is the same as the water, you sum the velocities of the river and the boat . 
I am Lyosha [343]3 years ago
8 0

Answer:Velocity of boat is 15.5 \frac{m}{s}

Explanation:Using equation of relative motion

V_{boat,water}=V_{boat}-V{water}

=>V_{boat}=V_{boat,water}+V_{water}

where velocity of boat relative to water , V_{boat,water}=18 m/s

velocity of water, V_{water}=-2.5\frac{m}{s}

and velocity of boat , V_{boat}=? \frac{m}{s} to be calculated.

Putting the known values in the above equation , we get

V_{boat}= (18-2.5)\frac{m}{s}=15.5\frac{m}{s}

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a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

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The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

#SPJ4

4 0
2 years ago
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