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3 years ago

10

On a time line.

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

Answer:

D

Explanation:

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dmitriy555 [2]

Answer:

3

Explanation:

it's too because on you measuring mass of something

6 0

3 years ago

Need help with question 1) and 2) please ASAP ??? 15 points ?

Vaselesa [24]

1- interaction between 2 objects
2- action- reaction force pairs

3 0

3 years ago

2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The tension is $T = 48.255 \ N$

b

The time taken is $t = 0.226 \ s$

c

The position for maximum velocity is

S = 0

d

The maximum velocity is $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

The mass of the bob is $m_b = 5 \ kg$

The angle is $\theta = 10^o$

The length of the string is $L = 0.5 \ m$

The tension on the string is mathematically represented as

$T = mg cos \theta$

substituting values

$T = 5 * 9.8 cos(10)$

$T = 48.255 \ N$

The motion of the bob is mathematically represented as

$S = A sin (w t + \frac{\pi }{2} )$

=> $S = A sin (wt)$

Where $w$ is the angular speed

and $\frac{\pi}{2}$ is the phase change

At initial position S = 0

So $wt = cos^{-1} (0)$

$wt = 1$

Generally $w$ can be mathematically represented as

$w = \frac{2 \pi }{T}$

Where T is the period of oscillation which i mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

So

$t = \frac{1}{w}$

$t = \frac{T}{2 \pi}$

$t = \sqrt{\frac{L}{g} }$

substituting values

$t = \sqrt{\frac{0.5}{9.8} }$

$t = 0.226 \ s$

Looking at the equation

$wt = 1$

We see that maximum velocity of the bob will be at S = 0

i. e $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

$V_{max} = w A$

Where A is the amplitude which is mathematically represented as

$A = L sin \theta$

So

$V_{max} = \frac{2 \pi }{T } L sin \theta$

$V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$ Recall $T = 2 \pi \sqrt{\frac{L}{g} }$

$V_{max} = \sqrt{gL} sin \theta$

substituting values

$V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

$V_{max} =0.384 \ m/s$

6 0

3 years ago

Modern scientists say that electrons are found in a ...

bulgar [2K]

I think it might be a or d i hope i have helped :)

8 0

3 years ago

Read 2 more answers

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago