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RideAnS [48]
3 years ago
7

A 30 kg child rides a 20 kg bicycle. Together, the child and the bicycle have a momentum of 110 kg-m/s. What is the velocity of

the boy and the bicycle?
Physics
1 answer:
love history [14]3 years ago
8 0

Answer:

The velocity of the boy and the bicycle is 2.2 m/s.

Explanation:

We have,

Mass of child is 30 kg and the mass of bicycle is 20 kg. The combined momentum of the child and the bicycle is 110 kg-m/s.

It is required to find the velocity of the boy and the bicycle. The momentum of an object is given in terms of mass and its velocity. So,

p=Mv

M is combined mass of child and bicycle

v=\dfrac{p}{M}\\\\v=\dfrac{110}{30+20}\\\\v=2.2\ m/s

So, the velocity of the boy and the bicycle is 2.2 m/s.

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ch4aika [34]

Incomplete question. However, I provided a brief about Kinetic energy generation.

<u>Explanation:</u>

Interestingly, Kinetic energy in simple terms refers to the energy possessed by a body in motion.

It is often calculated using the formula E =  \frac{1}{2}MV^{2}

A good example of creating even more kinetic energy is a hand crank toy car that moves after you wind it a little, when the car moves it is generating another measure of K.E.

7 0
3 years ago
A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi
Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

         \Delta E_{system} = E_{in} - E_{out}

        \Delta U = W_{in} - Q_{out}

Hence,    W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Therefore, we will calculate the final temperature as follows.

            \frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}

       T_{2} = \frac{20 psia}{14.7}(638 R)

                   = 868.03 R

Now, we will calculate the mass as follows.

             m = \frac{P_{1}V}{RT_{1}}

                 = \frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}

                 = 1.031 lbm

Hence,

        W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Putting the values into the above equation as follows.

            W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

    W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R

            W_{in} = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0
3 years ago
Atoms in Group 18 elements are inert (chemically unreactive) because ___________________________. A they combined to form molecu
Rom4ik [11]

When you say full valence shell, are you talking about a valence electron shell?

I am learning about atoms and i know a little bit

6 0
2 years ago
Read 2 more answers
The components of vector A are:
Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

tan\theta = \frac{opposite\: side}{adjacent\: side}

tan\theta = \frac{8.6}{6.1}

\theta = tan^{-1}1.41

\theta = 54.65^0

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

\theta = 360 - 54.65 = 305.3^0

so the correct answer will be 305 degree

3 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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