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3 years ago

A bus increases iys velocity from 20m/s to 30m/s in 2 seconds. Find ots acceleration

Physics

2 answers:

Katyanochek1 [597]3 years ago

4 0

Answer:

5ms^-2 be the acceleration

Vinvika [58]3 years ago

4 0

Explanation:

Solution,

Given,

u=20m/s

v=30m/s

t=2 seconds

Now,

According to the formula of acceleration,we have

a=v-u/t

=30-20/2

=10/2

=5m/s^2

So, it's acceleration is 5m/s^2

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Answer:

A. length, diameter, material, temperature

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2 years ago

A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/

klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

$d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m$

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

$\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}$

$-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}$ where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s t2 = 98.59s

So, the total time of flight will be:

$t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s$

6 0

3 years ago

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago

the maximum displacement of an oscillatory motion is A=0.49m. determine the position x at which the kinetic energy of the partic

kipiarov [429]

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is $\frac{1}{2}kA^2$

where k=spring constant

U+K.E.= $\frac{1}{2}kA^2$

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

$\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m$

3 0

3 years ago

The speed of an object in a set direction is called its what?

alisha [4.7K]

The speed of an object in a set direction is its velocity.

3 0

3 years ago

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A bus increases iys velocity from 20m/s to 30m/s in 2 seconds. Find ots acceleration​

A bus increases iys velocity from 20m/s to 30m/s in 2 seconds. Find ots acceleration