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3 years ago

5

2. Use the data table below to respond to the

1 answer:

forsale [732]3 years ago

8 0

Answer:

car d

Explanation:

5555555555555555555

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A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$

The potential energy reached by the ball at its rebound is

$U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J$

Thus the speed after the collision is

$\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}$

$v'=13.17\ m/s$

The change of momentum is

$\Delta P=0.0616\cdot (13.17+14)$

$\Delta P=1.67 \ kg.m/s$

(b) If the putty sticks to the floor, then v'=0

$\Delta P=0.0616\cdot (0+14)$

$\Delta P=0.86\ m/s$

3 0

4 years ago

A 1-lb block and a 100-lb block are placed side by side at the top of a frictionless hill. Each is given a very light tap to beg

qwelly [4]

Answer:

(c). The two blocks end in a tie

Explanation:

the reason being the absence of any resistance offered to both of the blocks.

if the slope of the hill is for instance 60 deg.

then the acceleration in absence of any resistance is a= 9.81sin(60)

since the acceleration is same then both of the blocks will reach at the same instant

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4 years ago

A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have a

Verizon [17]

Answer:

The shear deformation is $\Delta x=3.34\times10^{-6}\ m$ .

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus $S = 1\times10^{9}\ N/m^2$

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

$S=\dfrac{Fl_{0}}{A\Delta x}$

$\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}$

$\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}$

Put the value into the formula

$\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}$

$\Delta x=3.34\times10^{-6}\ m$

Hence, The shear deformation is $\Delta x=3.34\times10^{-6}\ m$ .

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Which of the following was not a factor that got the public interested in NASA? A. real people B. scientific reports C. adventur

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B. Scientific reports

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What mass of electrons would be required to just neutralize the charge of 5.0 g of protons? (the mass of a proton is 1.67262×10−

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<span>To neutralize the charge of 1.0g of electrons?</span>

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