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2 years ago

How do the weights of the tug of war teams affect the match

2 answers:

8 0

In Newton's Second Law of Motion, it simply states that mass x acceleration= force. So if there's more weight on one side and the same amount of acceleration on both side, then the side with more weight has more force. And with more force it tugs it more than the other side.

Shkiper50 [21]2 years ago

3 0

Because when the players lean towards their side, gravity pulls them backwards, it also makes it harder for the other side to pull someone so heavy. It's like a kid trying to pull their parent. Their parent is so much heavier that if the parent doesn't want to move they won't.

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When a person plucks a guitar string, the number of half wavelengths that fit into the length of the string determines the _____

Vedmedyk [2.9K]

Answer:

Frequency

Explanation:

Each half wavelength has a point of largest amplitude (aka a node). Depending on the wavelength each node oscillates at a certain rate of swings per unit of time. The latter is referred to as frequency and measure in Hertz [Hz].

8 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

Which atmospheric condition is most likely responsible for the wind blowing the clouds from the sea towards the land?

lions [1.4K]

Uneven heating of land and sea causes warm air over land to rise up, creating a low pressure zone. So wind blows in from the sea to fill this low pressure zone

3 0

2 years ago

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of

Harman [31]

Answer:

4 smaller disks

Explanation:

We are given;

Mass of smaller and larger disks = M

Radius of smaller disk = R

Radius of larger disk = 4R

Formula for moment of inertia about cylinder axis is:

I = ½MR²

Thus;

For small disk, I_small = ½MR²

For large disk, I_large = ½M(2R)² = 2MR²

We are told that moment of inertia of System A consists of two of the larger disks. Thus;

I_A = 2 × I_large = 2 × 2MR²

I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

I_A = I_B

So;

4MR² = MR²(2 + n/2)

MR² will cancel out to give;

4 = 2 + n/2

Multiply through by 2 to give;

8 = 4 + n

n = 8 - 4

n = 4

5 0

3 years ago

It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo

vovikov84 [41]

Answer:

L = 1.11 x $10^{6}$ m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x $10^{5}$ m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x $10^{-2}$ m

Length = ?

So,

we know that,

U = 1/2 C Δ $v^{2}$

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ $v^{2}$

52000 J = (0.5) x (C) x ( $610^{2}$ )

C = 0.28 F

And we also know that,

C = $\frac{K*E*A}{d}$

E = 8.85 x $10 ^{-12}$

K = 3.7

A = 0.20 x L

d = 2.6 x $10^{5}$ m

Plugging in the values into the formula, we get:

0.28 = $\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }$

Solving for L, we get:

L = 1.11 x $10^{6}$ m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

7 0

3 years ago