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3 years ago

6

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

its final position?

1 answer:

ycow [4]3 years ago

5 0

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

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3. If a car is moving at 90km/hr and it rounds a corner, also at 90km/hr. Does it maintain

dolphi86 [110]

Answer:

Constant speed: yes

Constant velocity: no

Explanation:

Let's remind the definition of speed and velocity:

- Speed is a scalar quantity, which is equal to the ratio between the distance covered (regardless of the direction) and the time taken:

$s=\frac{d}{t}$

- Velocity is a vector quantity, so it has both a magnitude and a direction. The magnitude is equal to the rate between the displacement of the object and the time taken, while the direction is the same as the displacement.

In this problem, we notice that:

- The speed of the car remains constant, as it is 90 km/h

- However, its direction of motion changes while the car travels round the corner: this means that the direction of the velocity is also changing, therefore velocity is not constant.

8 0

4 years ago

A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of

Wewaii [24]

(1) The wavelength of the wave is 1.164 m.

(2) The velocity of the wave is 23.7 m/s.

(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

<h3> Wavelength of the wave</h3>

A general wave equation is given as;

y(x, t) = A sin(Kx - ωt)

<h3>Velocity of the wave</h3>

v = ω/K

From the given wave equation, we have,

y(x, t) = 0.048 sin(5.4x - 128t)

v = ω/K

where;

ω corresponds to 128
k corresponds to 5.4

v = 128/5.4

v = 23.7 m/s

<h3>Wavelength of the wave</h3>

λ = 2π/K

λ = (2π)/(5.4)

λ = 1.164 m

<h3>Maximum speed of the wave</h3>

v(max) = Aω

where;

A is amplitude of the wave
ω is angular speed of the wave

v(max) = (0.048)(128)

v(max) = 6.14 m/s

Thus, the wavelength of the wave is 1.164 m.

The velocity of the wave is 23.7 m/s.

The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

Learn more about wavelength here: brainly.com/question/10728818

#SPJ1

3 0

2 years ago

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

b)

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

3 years ago