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2 years ago

Why a paperclip does NOT float?

2 answers:

4 0

Answer: the tention of the water since the object is light the water particles can rise it so it's not entilely floating

Explanation: flat line:_____________________________

Nataliya [291]2 years ago

3 0

Because of Surface tension

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As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

poizon [28]

Answer:

$4.048\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

E = Energy = $4.91\times 10^{-19}\ J$

Wavelength ejected is given by

$\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m$

The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is $4.048\times 10^{-7}\ m$

4 0

3 years ago

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago

In the year 2055, a rocket was launched from a research laboratory on Mars. Mars has essentially no atmosphere. The test rocket

algol [13]

Answer:

h = 618.64 m

Explanation:

First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:

h₁ = Vit + (1/2)at²

where,

h₁ = height gained during the burning of fuel

Vi = Initial Velocity = 0 m/s

t = time = 7 s

a = acceleration = 8 m/s²

Therefore,

h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²

h₁ = 196 m

Now we use 1st equation of motion to find final speed Vf:

Vf = Vi + at

Vf = 0 m/s + (8 m/s²)(7 s)

Vf = 56 m/s

Now, we calculate height covered in free fall motion. Using 3rd equation of motion:

2ah₂ = Vf² - Vi²

where,

a = - 3.71 m/s²

h₂ = height gained during free fall motion = ?

Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)

Vi = 56 m/s

Therefore,

(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²

h₂ = 422.64 m

So the total height gained will be:

h = h₁ + h₂

h = 196 m + 422.64 m

4 0

2 years ago

A 43-kg child sits in a massless swing. With what horizontal force must the seat be pulled so that the ropes form an angle of 35

GenaCL600 [577]

Answer:

F_x = 295.4N

Explanation:

180°-90°-35°= 55°

F= m(a)

F_y = 43kg(9.81m/s²)

F_y = 421.83N

tan(55°) = 421.83N/F_x

F_x = 421.83N/tan(55°)

F_x = 295.3685458N

7 0

2 years ago

Will the string broke and why

GaryK [48]

It will most likely break. The frame is 5 newton's and the string can only hold 2.5 newton's.

5 0

3 years ago