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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

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

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

$T = 190 N$

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A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.

Jobisdone [24]

Answer:

<em>(a) t = 4.52 sec</em>

<em>(b) X = 1,156.49 m</em>

Explanation:

<u>Horizontal Launching </u>

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity $v_o$ remains the same in time. The distance x is computed as .

$\displaystyle x=v_o.t$

(a)

The vertical component of the velocity $v_y$ starts from zero and gradually starts to increase due to the acceleration of gravity as follows

$v_y=gt$

This means the vertical height is computed by

$\displaystyle h=h_o-\frac{gt^2}{2}$

Where $h_o$ is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

$\displaystyle t=\sqrt{\frac{2h_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec$

(b)

We now compute the horizontal distance knowing $v_o=256\ m/s$

$\displaystyle x=(256).(4.52)$

$X=1,156.49\ m$

4 0

2 years ago

Before the widespread use of computers, how was the epicenter of an earthquake determined?

Nezavi [6.7K]

Data was used from three different stations. Where their data overlapped was where the epicenter was located.
it is called triangulation

7 0

2 years ago

Read 2 more answers

The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$

Since the definition of the base logarithm 10 of 10 is equal to 1 then

$6*1=6$

The value of the given logarithm is equal to 6

8 0

3 years ago

A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

meriva

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

8 0

3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago