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2 years ago

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A 0.180-kilogram cart traveling at 0.80 meter per second to the right collides with a 0.100-kilogram cart initially at rest. The

carts lock together upon collision. Calculate the final velocity of the carts.

1 answer:

My name is Ann [436]2 years ago

3 0

According to the <u>saving of the momentum</u> law, the total momentum <em>before </em>collision equals to the total momentum <em>after </em>collision

After collection, the 2 carts lock together, one body with a mass $m=m_{1}+m_{2}$ and velocity $v$

$(P_{1}+P_{2})_{before}=P_{after}$

from the definition of the momentum $P=mv$

$P_{1,before}=m_{1}v_{1,before}=0.180*0.80=0.144kg.m/s$

$P_{2,before}=m_{2}v_{1,after}=0.100*0=0kg.m/s$

thus

$P_{after}=(0.144+0)_{before}=0.144 kg.m/s$

we calculate the velocity

$v=\frac{P}{m}=\frac{0.144}{0.180+0.100}= 0.514m/s$

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A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

8 0

2 years ago

A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots

Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

$L_i = L_f$

$mv \frac{L}{2} = (I_1 + I_2)\omega$

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

$I_1 = \frac{ML^2}{3}$

$I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2$

$I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2$

now from above equation

$0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega$

$\omega = 5.85 rad/s$

8 0

2 years ago

Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms

MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = $6 *10^{-3} seconds$

So acceleration of frog = $\frac{3.7}{60*10^{-3}}$

= 61.66 m/ $s^2$

So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

8 0

2 years ago

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A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. How much weight can be placed on the

Setler79 [48]

The answer is 96 N .....................................

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2 years ago

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