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Stels [109]
3 years ago
12

A biker pedals with a constant acceleration to reach a velocity of 7.5m/s over 4.5s. During the period of acceleration, the bike

's displacement is 19m. What was the initial velocity of the bike?
Physics
1 answer:
Flura [38]3 years ago
7 0

Given:

v(final velocity)= 7.5 m/s

t(time taken to pedal along)=4.5 s

Displacement (s)=19 m

Now we know that

s=ut+1/2(at^2)

Where s is the displacement measured in m

u is the initial velocity measured in m/sec

a is the acceleration measured in m/s^2.

t is the time taken to cover this distance.

Substituting the given values in the above formula we get

19= 4.5u+1/2(a x 4.5 x 4.5)

20.25 a + 9 u = 38


Now we also know that

v= u + at

Substituting the given values in the above formula we get

7.5= u + 4.5a


Solving for u and a from the above equations we get

u = 0.944m/s

a= 1.45 m/s^2

Thus the initial velocity is 0.944 m/s


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A liquid with a density of 900kg/m 3 is stored in a pressurized, closed storage tank. The tank is cylindrical with a 10m diamete
Orlov [11]

Answer:

18.62 m/s

Explanation:

Given that:

A liquid with a density of 900 kg/m 3 is stored in a pressurized, closed storage tank.

Diameter of the tank = 10 m

The absolute pressure in the tank above the liquid is 200 kPa = 200, 000 Pa

At pressure of 200 kPa   ; the final velocity = 0

Atmospheric pressure at 5cm = 101325 Pa

We are to calculate the initial velocity of a fluid jet when a 5cm diameter orifice is opened at point A?

By using Bernoulli's theorem between the shaded portion in the diagram;

we have:

Pa \ + \ \frac{1}{2} \  \delta \ v^2_1 \ + \ \delta gy_1 = P + \delta gy_2

\frac{1}{2} \  \delta \ v^2_1 \ = P + \delta gy_2 -  \ \delta gy_1  - Pa

\frac{1}{2} \  \delta \ v^2_1 \ =  \delta g(y_2 -y_1 )+   ( P   - Pa )

v_1 \ =  \sqrt{ \frac {2 \ \delta g(y_2 -y_1 )+  2  ( P   - Pa )}{\delta}}

where;

Pa = atmospheric pressure = 101325 Pa

\delta = density of liquid = 900 kg/m³

v_1 = initial velocity = ???

g = 9.8 m/s²

y_1 = height of the hole from the buttom

y_2 = height of the liquid surface from the button

v_1 \ =  \sqrt{ \frac {2*900*9.8(7 -  0.5 )+  2  ( 200,000   - 101325 )}{900}}

v_1 = 18.62 \ m/s

Thus, the initial  velocity of the fluid jet  = 18.62 m/s

3 0
3 years ago
You observe two stars over the course of a year (or more) and find that both stars have measurable parallax. (1 arc second is 1/
Ierofanga [76]

Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

The distance between the Earth and the Sun is 150000000 Km. That distance is also known as an astronomical unit (1AU).

The parallax angle can be defined in the following way:

\tan{p} = \frac{1AU}{d}    

Where d is the distance to the star.

p('') = \frac{1}{d}    (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

d(pc) = \frac{1}{1}

d(pc) = 1 pc

So, star X is at 1 parsec from Earth.

Case of Star Y (p('') = \frac{1}{2}):

d(pc) = \frac{1}{(\frac{1}{2})}

d(pc) = 2 pc

So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

5 0
3 years ago
A plane mirror is placed to the right of an object. The image formed by the mirror will be a
guajiro [1.7K]

Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

3 0
3 years ago
Read 2 more answers
A car goes from rest to a speed of 90 km/h in 10 seconds . What is the car's acceleration in m/s²
zloy xaker [14]
First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s

Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!
7 0
3 years ago
Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.
LiRa [457]

Answer:

         Time - taken = 2.5 s

          deceleration= -8 m/s²

Solution:

            Given:

                     speed, v = 8 m/s

                 distance, d = 20m

                     

              To Find:

                     deacceleration = ?

               

               As we know speed is defined as

                          v = d/t

                plugging in the values

                          t =  20/ 8

                          t = 2.5s

                Now from deceleration formula

                        a =  - v/ t

                        a = - 20/ 2.5

                        a = - 8 m/s²

          Thus, the time taken and acceleration is 2.5 s and -8 m/s²

          respectively.

          Learn more about deceleration here:

                brainly.com/question/13354629

                       #SPJ4  

               

                       

             

7 0
2 years ago
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