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2 years ago

4 WILL GIVE YOU BRAINLIST

Physics

2 answers:

Arisa [49]2 years ago

5 0

The last one, handmade gifts require more of the givers time!

salantis [7]2 years ago

5 0

The claim is looking for a statement that supports why handmade gifts are more meaningful.

So the answer is . . .

Handmade gifts require more of the giver’s time and effort than store-bought gifts.

This is the correct answer because putting effort and time into a gift to make it original and one of a kind makes a handmade gift more meaningful than going to the store finding a gift.

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Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down

VashaNatasha [74]

U can always just do the classic roller coaster going up an incline and create some sort of story from that.

8 0

3 years ago

What travels with a sound wave matter, energy or both?

il63 [147K]

Energy travels with sound waves.

8 0

3 years ago

Read 2 more answers

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago

A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the r

kipiarov [429]

Answer:

The torque exerted on the merry-go-round is 766.95 Nm

Explanation:

Given;

mass of the merry-go-round, m = 416 kg

radius of the disk, r = 1.7 m

angular speed of the merry-go-round, ω = 3.7 rad/s

time of motion, t = 2.9 s

The torque exerted on the merry-go-round is calculated as;

$\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm$

Therefore, the torque exerted on the merry-go-round is 766.95 Nm

6 0

2 years ago

As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.01 mm apart and position

Yakvenalex [24]

Answer:

$0.00195\ \text{m}$

$0.00293\ \text{m}$

Explanation:

m = Order = 1

D = Distance between screen and slit = 3.09 m

d = Slit distance = 1.01 mm

$\lambda$ = Wavelength = 639 nm

Distance from the first bright fringe from the central bright fringe is given by

$y=\dfrac{m\lambda D}{d}\\\Rightarrow y=\dfrac{1\times 639\times 10^{-9}\times 3.09}{1.01\times 10^{-3}}\\\Rightarrow y=0.00195\ \text{m}$

Distance from the first bright fringe from the central bright fringe is $0.00195\ \text{m}$

Distance from the second dark fringe from the central bright fringe is given by

$y=(m+\dfrac{1}{2})\dfrac{\lambda D}{d}\\\Rightarrow y=(1+\dfrac{1}{2})\dfrac{639\times 10^{-9}\times 3.09}{1.01\times 10^{-3}}\\\Rightarrow y=0.00293\ \text{m}$

Distance from the second dark fringe from the central bright fringe is $0.00293\ \text{m}$ .

8 0

2 years ago