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ch4aika [34]
3 years ago
13

Circuits

Physics
1 answer:
marusya05 [52]3 years ago
7 0

Answer:

ഹലോ എങ്ങനെയുണ്ട് ക്ഷമിക്കണം എനിക്ക് ഉത്തരം അറിയില്ല

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The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a wedge placed under the 40 cm mark and car
Nitella [24]

Answer:

W = 30 N

Explanation:

Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.

(70\ N)(40\ cm - 10\ cm)-(30\ N)(50\ cm-40\ cm)-(W)(100\ cm - 40\ cm) = 0\\W(60\ cm) = (70\ N)(30\ cm)-(30\ N)(10\ cm)\\\\W = \frac{1800\ N.cm}{60\ cm}

<u>W = 30 N</u>

6 0
3 years ago
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
den301095 [7]

Answer:

\theta=7^o

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

\vec r_1=\ km

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

\vec r_2=\ km=\ km

Finally, he drives 9.5 km 35° north of east.

\vec r_3=\ km=\ km

The total displacement is

\vec r_t=\ km+\ km+\ km

\vec r_t=\ km

The direction can be calculated with

\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232

\boxed{\theta=7^o}

7 0
3 years ago
The cannon on a battleship can fire a shell a maximum distance of 26.0 km.
Alla [95]
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is \theta = 45^{\circ}.
In fact, the laws of motions on both x- and y- directions are
S_x(t)= v_0 cos \theta t
S_y(t)= v_0 \sin \theta t -  \frac{1}{2} gt^2
From the second equation, we get the time t at which the projectile hits the ground, by requiring S_y(t)=0, and we get:
t= \frac{2 v_0 \sin \theta}{g}
And inserting this value into Sx(t), we find
S_x(t) = 2  \frac{v_0^2}{g}  \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)
And this value is maximum when \theta=45^{\circ}, so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
S_x(t)=  \frac{v_0^2}{g} \sin (2\theta)
And by using S_x = 26 km=26000 m, we find the value of the initial velocity v0:
v_0 =  \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s
8 0
3 years ago
What will happen to the momentum of the car as sand is loaded onto it?
ivanzaharov [21]
It seems reasonable to assume that the car is standing still while the sand is pouring in. Since its speed is zero, it has no momentum, and that won't change until it starts moving, no matter how much sand has been loaded onto it.
4 0
3 years ago
what is the period of a sound wave whose wavelength is 20.0m? use values from the book and show all of your work
Lera25 [3.4K]
The wavelength of a sound wave is related to its frequency by the relationship:
f= \frac{v}{\lambda}
where
f is the frequency
v is the speed of the wave
\lambda is the wavelength

The wave in our problem has wavelength of \lambda=20.0 m and speed of v=343 m/s (this is the speed of sound in air), therefore its frequency is
f= \frac{343 m/s}{20.0 m}=17.15 Hz

And the period of the wave is equal to the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{17.15 Hz}=0.058 s
5 0
4 years ago
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