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2 years ago

10

Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.

Make a hypothesis: Who do you think would be slowed down more by frictional forces on the ice, the small child or the large adult?

2 answers:

kramer2 years ago

8 0

You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.

Aleksandr [31]2 years ago

4 0

The larger father, the more weight the more friction. Hope this helped.

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Henri draws a wave that has a 4 cm distance between the midpoint and the trough. Geri draws a wave that has an 8 cm vertical dis

andre [41]

Answer:

Henri’s wave and Geri’s wave have the same amplitude and the same energy

Explanation:

The amplitude of a wave is the distance between the midpoint and the trough (or the crest). This is equivalent to half the distance between the trough and the crest. Therefore:

amplitude of Henri's wave: 4 cm

amplitude of Geri's wave: 8/2 = 4 cm

The energy of a wave is directly proportional to its amplitude.

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2 years ago

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The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

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2 years ago

If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

GenaCL600 [577]

Answer:

I am not a driver, but I think it's C.

Explanation:

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2 years ago

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A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei

QveST [7]

gravitational potential energy is given by formula

$U = mgh$

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

$\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}$

$\frac{U_2}{U_1} =\frac{m_2}{m_1}$

given that

$m_2 = 4 m_1$

now we have

$\frac{U_2}{U_1} = 4$

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2 years ago