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2 years ago

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Two identical 3kg uniform rods are hinged at point b and attached to a spring of constant k = 4kn/m. the rods are guided by smal

l wheels of negligible mass and the system is in equilibrium when the rods are horizontal as shown. knowing that point b is depressed 20 mm and rele

1 answer:

Gnesinka [82]2 years ago

5 0

Thank for using Brainly. The answer is 15mm

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For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to

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Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

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2 years ago

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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

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2 years ago

(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

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3 is jxhuneuxndnzixbf

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The table shows information about four students who are running around a track. Which statement is supported by the information

Vikentia [17]

Answer:

<em>Correct option: Mohammed has less kinetic energy than Autumn.</em>

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its motion. If the object has a mass m and travels at a speed v, then the kinetic energy K is:

$\displaystyle K=\frac{1}{2}mv^2$

The information about four students includes their mass and velocity as follows:

Autumn has a mass of m1=50 kg and a velocity (magnitude) of v1=4 m/s, thus their kinetic energy is:

$\displaystyle K_1=\frac{1}{2}50\cdot 4^2$

$K_1=400\ J$

Mohammed has a mass of m2=57 kg and a velocity (magnitude) of v2=3 m/s, thus their kinetic energy is:

$\displaystyle K_2=\frac{1}{2}57\cdot 3^2$

$K_2=256.5\ J$

Lexy has a mass of m3=53 kg and a velocity (magnitude) of v3=2 m/s, thus their kinetic energy is:

$\displaystyle K_3=\frac{1}{2}53\cdot 2^2$

$K_3=106\ J$

Chiang has a mass of m4=64 kg and a velocity (magnitude) of v4=5 m/s, thus their kinetic energy is:

$\displaystyle K_4=\frac{1}{2}64\cdot 5^2$

$K_4=800\ J$

Sorted from lower kinetic energy to higher:

Lexy, Mohammed, Autumn, Chiang. Thus:

Autumn has more kinetic energy than Chiang. False

Mohammed has less kinetic energy than Autumn. True

Lexy has more kinetic energy than Mohammed. False

Chiang has less kinetic energy than Lexy. False

Correct option: Mohammed has less kinetic energy than Autumn.

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2 years ago

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8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

hence, B_o = 1.013μT

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2 years ago