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3 years ago

A mass weight of 120N is hung from two strings. what is the tension?

Physics

2 answers:

I am Lyosha [343]3 years ago

8 0

Answer: $T_1 = 77.2 \text N \newline T_2= 91.87 \text N$

Assuming the figure as attached.

The sum of Horizontal components of the Tension in the rope is zero or equal as they balance out each other. $T_1 sin40 \textdegree = T_2 sin50 \textdegree \newline T_1 = 1.19 T_2$

The sum of vertical components of the tension is equal to the weight of the block. $T_1 cos 40 \textdegree +T_2cos50 \textdegree = 120 \text N \newline 1.19T_2(0.766 )+T_2 0.64 = 120 \text N T_2 = 77.2 \text N T_1 = 1.19 T2 = 1.19 \times 77.2 \text N = 91.87 \text N$

kramer3 years ago

7 0

The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

PLZ Help Me I give you brainlist also NO LONKS OR I REPORT and Have a gr8 day my peeps

Sergio039 [100]

Answer:

the answer is a because I saw it in a syllabus

3 0

3 years ago

A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

Musya8 [376]

Answer:

162500000.

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

$n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}$

n=162500000 t

$\dfrac{n}{t}=16250000$

The number of electron passe per second will be 162500000.

4 0

3 years ago

20. A semiconductor is a

Stells [14]

Answer:

D. crystalline solid that conducts current under certain conditions

Explanation:

Semiconductors are crystalline solids that has the ability to conduct electrical currents but on certain conditions e.g heat. The conduction of semiconductors is less than that of conductors (metals) but more than insulators (nonmetals), hence, they are said to be intermediates of conductors and insulators in terms of electrical conductivity.

Examples of semiconductors are silicon, boron, carbon, germanium, arsenic etc.

5 0

3 years ago

One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6

Angelina_Jolie [31]

Solution :

Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.

It is given that :

Successive harmonic frequencies, f = 52.2 Hz

and f' = 60.9 Hz

Therefore, fundamental frequency, F = f' - f

F = 60.9 - 52.2

F = 8.7 Hz

Therefore the string which is fixed at both the ends forms all the harmonics.

6 0

2 years ago