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german
3 years ago
12

Stu wanted to calculate the resistance of a light bulb connected to a 4.0-V battery, with a resulting current of 0.5 A. He used

the formula R = VI and obtained an answer of 2 . Was Stu’s answer correct? How do you know?
Physics
2 answers:
Alexxandr [17]3 years ago
7 0

Answer:

<u><em>Sample Response</em></u>: Stu's answer was incorrect. According to Ohm's law, R = V/I. This is the equation that should be used to obtain the correct answer, which is 8 ohms.

Explanation:

Straight copied and pasted from edge. 2020

Margarita [4]3 years ago
6 0

Answer:

The answer is not correct.

Explanation:

Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.

V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\

In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.

R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]

That is the reason that the result found by Stu is not correct.

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho
Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

F_{down} L_1 = W_{snow} L_2

so here we have

F_{down} = \frac{L_2}{L_1} (W_{snow})

so here we can say that upward force by which we push up is always more than the downwards force

8 0
3 years ago
How do scientists decide how they gather information
kogti [31]
The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
3 0
3 years ago
A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.
Alex73 [517]

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

                                 d' = d cos Ф

                                     = 1430 cos 5.76°

                                     = 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

5 0
3 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
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