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stich3 [128]
3 years ago
10

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

e string. if each washer has a mass of 12 g, what is the acceleration of the system, in m/s2
Physics
1 answer:
svetlana [45]3 years ago
5 0
Let's call m=565~g=0.565~kg the mass of the glider and m_w=7\cdot12~g =84~g=0.084~kg the total mass of the seven washers hanging from the string. 
The net force on the system is given by the weight of the hanging washers:
F_{net} = m_w g
For Newton's second law, this net force is equal to the product between the total mass of the system (which is m+m_w) and the acceleration a:
F_{net}=(m+m_w)a
So, if we equalize the two equations, we get
m_w g = (m+m_w)a
and from this we can find the acceleration:
a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2
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Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
Andreas93 [3]

Answer:

The displacement is  \Delta H =    -  1 \ m

The distance  is  D =  4 \  m

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  h  =  1 \ m

    The height attained at  the first bounce is  h_1  = 0.8  \  m

    The height attained at  the second bounce is   h_2 = 0.5 \  m

    The height attained at  the third bounce is h_3 = 0.2 \  m

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            \Delta H =  (0  -  h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           \Delta H =  (0  -  1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)

=>          \Delta H =    -  1 \ m

Generally the distance covered by the ball is mathematically represented as  

                D =  h +  2h_2 + 2h_3 + 2h_3

The 2 shows that the ball traveled the height two times

              D =  1 +  2* 0.8  + 2* 0.5 + 2* 0.2

=>           D =  4 \  m

     

5 0
3 years ago
A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through th
GenaCL600 [577]

Answer:

\rho=1.8\times 10^{-8}\Omega.m

Explanation:

Given:

width of the wire, w=1.8\ mm=1.8\times 10^{-3}\ m

thickness of the flat wire, d=0.12\ mm=1.2\times 10^{-4}

length of the wire, l=12\ m

voltage across the wire, V=12\ V

current through the wire, I=12\ A

Now the net resistance of the wire:

using ohm's law

R=\frac{V}{I}

R=\frac{12}{12}

R=1\ \Omega

We have the relation between the resistivity and the resistance as:

R=\rho.\frac{l}{a}

where:

a = cross sectional area of the wire

\rho = resistivity of the wire material

1=\rho\times \frac{12}{1.8\times 10^{-3}\times 1.2\times 10^{-4}}

\rho=1.8\times 10^{-8}\Omega.m

4 0
3 years ago
A surfer is actually traveling on which part of a surface wave
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A surfer travels on the crest of a surface wave
8 0
3 years ago
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If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno
vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0
3 years ago
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Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

7 0
3 years ago
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