Ask question

Ask question

All categories

ICE Princess25 [194]

2 years ago

11

A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100 N. The magnitude of the tension f

orce of each of the two supporting ropes is A : 190 N B : 120 N C : 60 N D : 94 N

1 answer:

guajiro [1.7K]2 years ago

3 0

Answer:

T = 190 N

Explanation:

When child is sitting on the swing then the weight of the child is vertically downwards

So it is

$F_g = 160 N$

now a force of 100 N is acting on the swing in horizontal direction

so it is given as

$F_x = 100 N$

now the net force is resultant force due to gravity and horizontal force

so it is given as

$T = \sqrt{F_x^2 + F_g^2}$

$T = \sqrt{100^2 + 160^2}$

$T = 190 N$

You might be interested in

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

Swings are usually made out of materials that do not conduct heat easily, so that children will not burn themselves as they play

Irina18 [472]

the correct answer is D: all of these

8 0

3 years ago

Describe the sequence of mechanical energy events that lets you hear the

lina2011 [118]

Answer:

Starting from the beginning.

There is a radio signal that is received by the radio.

The radio interprets the signal and produces a current in response to it.

That current goes to a membrane that oscillates producing sound, the oscillation of the membrane is the first mechanical energy event here.

These oscillations can travel in material mediums, for example, the air. Then there is a production of waves (soundwaves) that travel in the air (second event).

Those waves now hit the wall that separates you and your neighbor, as the wall is made of a material, the soundwaves can travel through it, but they will be dispersed (a part of the waves rebounds on the wall, and another part is dissipated as the wave travels through the wall), there is also a transmitted part of the wave, that is now in your house. (this change of medium will be the third event). Now only the lower frequencies survive, this is why the sound is "muffled".

Those remaining frequencies now travel in your house, and when they reach your ear, your ear sends a signal to your brain and your brain interprets them as sound. The wave interacting with your ear will be the fourth and last mechanical energy event.

3 0

2 years ago

How many genders are there?

Dvinal [7]

Answer:2

Explanation:

8 0

3 years ago

Read 2 more answers

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago