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2 years ago

Conclusion for ohm law

1 answer:

7 0

Ohm’s Law states that electrical current is proportional to voltage and therefore inversely proportional to resistance - V = I x R

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3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.

gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

K.E = ?

According to conservative of energy

K.E = P.E

K.E = mgh

K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

6 0

3 years ago

Atomic math challenge will give brainly and thanks

nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794 ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0

3 years ago

Read 2 more answers

when white light is incident on prism, which one of the resulting color components will have the lowest index of refraction?

attashe74 [19]

The color components that will have the lowest index of refraction will be orange.

3 0

2 years ago

Read 2 more answers

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers

g Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an an

Mamont248 [21]

Answer:

add the angles

HOPE you understand

5 0

2 years ago