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3 years ago

3. A roller coaster car sits at rest at the top of a hill, 23.5 meters tall. What will be the final

Physics

1 answer:

Sloan [31]3 years ago

5 0

Answer:

21.5 m/s

Explanation:

KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 23.5 m)

v = 21.5 m/s

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9. A soft drink distibutor drove west 11 km from Minnehaha to Nut Plains. She then turned around and drove back east for 209 km

guajiro [1.7K]

Answer:

220km

Explanation:

Given parameters:

Distance traveled from Minnehaha to Nut Plains = 11km

Distance traveled to Dover = 209km

Unknown:

The distance the soft drink distributor traveled = ?

Solution:

To solve this problem, we must understand that distance is the total length of path covered in a journey.

So;

Distance = 209km + 11km = 220km

The soft drink distributor covered 220km

3 0

2 years ago

A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.

MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm= $\frac{20}{100}=0.20m$

1 m=100 cm

Mass of rod,M=190 g= $\frac{190}{1000}=0.19kg$

Mass of ball,m=19 g= $\frac{19}{1000}=0.019 kg$

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

$I_{rod-clay}=I_{rod}+I_{clay}$

$I_{rod-clay}=\frac{1}{3}ML^2+mL^2$

Substitute the values then we get

$I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2$

$I_{rod-clay}=3.29\times 10^{-3} kgm^2$

Now, the center of mass of the combination of the rod and clay is given by

$y=\frac{Md_1+md_2}{M+m}$

Substitute $d_1=\frac{L}{2}$ =Distance between pivot and the center of the rod

$d_2=L=$ The distance between rod and clay

Using the formula

$y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}$

$y=0.1091$ m

Time period of the oscillation of the system of the rod and the clay is given by

$T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}$

g= $9.8m/s^2$

Using the formula

Time period= $2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}$

Time-period=0.76 s

Hence, the period =0.76 s

8 0

2 years ago

Calculate the moment of inertia of a skater given the following information. The skater with arms extended is approximately a cy

Bad White [126]

Answer:

The moment of inertia of a skater is : $I= 2.343 kgm^{2}$

Explanation:

we can use parallel theorem to find the moment of inertia of a skater.

Arm extended cylinder mass: $M= 52.5 kg$

Radius: $R= 0.110 m$

Length: $L= 0.900 m$

and $m=3.75 kg$

$I=\frac{1}{2} MR^{2} +[\frac{1}{2} mL^{2} +m(R+\frac{L}{2} )^{2} ]$

$I=\frac{1}{2} (52.5)(0.110)^{2} +[\frac{1}{12} (3.75)(0.900)^{2} +3.75(0.110+\frac{0.900}{2} )^{2} ]$

$I=2.343 kgm^{2}$

4 0

2 years ago

A piece of metal has a density of 11.3 g/cm and a volume of

Juli2301 [7.4K]

Answer:75.71g≈75.7

Explanation:Mass = Density ⋅ Volume = 11.3 ⋅ 6.7 = 75.71g

3 0

2 years ago

Convert 26.4 mi to km

aev [14]

Answer:

42.48 [km]

Explanation:

To convert miles to kilometers we must know the conversion factor which is equal to:

$1.609[\frac{km}{mi} ]$

We make the product of the value required by the conversion factor:

$26.4[mi]*1.609[\frac{km}{mi} ]\\\\$

26.4*1.609 = 42.48 [km]

6 0

2 years ago