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3 years ago

A flashlight bulb is connected to a dry cell of voltage 2.25 V. It draws 35.0 mA (1000 mA = 1 A). What is its resistance?

Physics

1 answer:

Anarel [89]3 years ago

6 0

The answer & explanation for this question is given in the attachment below.

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A 30.0-kgkg box is being pulled across a carpeted floor by a horizontal force of 230 NN , against a friction force of 210 NN . W

Agata [3.3K]

Answer:

The acceleration of the box is 0.67 m/s²

Explanation:

Given that,

Mass of box = 30.0 kg

Horizontal force = 230 N

Friction force = 210 N

We need to calculate the acceleration of the box

Using balance equation

$F-f_{k}=ma$

$a=\dfrac{F-f_{k}}{m}$

Where, F = horizontal force

$f_{k}$ =frictional force

m= mass of box

a = acceleration

Put the value into the formula

$a=\dfrac{230-210}{30}$

$a=0.67\ m/s^2$

Hence, The acceleration of the box is 0.67 m/s²

4 0

3 years ago

Rita places a 2.5 kg block on a frictionless inclined plane that is 30 degrees above horizontal. She applies a horizontal force,

vitfil [10]

Answer:

14.2

Explanation:

find horizontal force of the weight = 2.5kg x 9.8 Sin30 = 12.3 N

to prevent the sliding she needs to pull horizontally

Fh = 12.3 /Cos 30 =14.2N

7 0

3 years ago

Can you help me answer this?

Pavel [41]

The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.

5 0

4 years ago

A 70-cm-diameter wheel accelerates uniformly from 160rpm to 280rpm in 4.0s. Determine (a) its angular acceleration, and (b) the

poizon [28]

Explanation:

Given that,

Diameter of the wheel, d = 70 cm = 0.7 m

Initial angular speed, $\omega_i=160\ rpm=16.75\ rad/s$

Final angular speed, $\omega_f=280\ rpm=29.32\ rad/s$

Time, t = 4 s

(a) Angular acceleration,

$\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{29.32-16.75}{4}\\\\\alpha =3.14\ rad/s^2$

(b) Tangential acceleration is :

$a=r\alpha \\\\a=0.35\times 3.14\\\\a=1.085\ m/s^2$

Angular speed of the wheel after 2 seconds is :

$\omega_f=\omega_i+\alpha t\\\\\omega_f=16.75+3.14\times 2\\\\\omega_f=23.03\ rad/s$

Radial acceleration will be :

$a=\omega_f^2r\\\\a=(23.03)^2\times 0.35\\\\a=185.6\ m/s^2$

Hence, this is the required solution.

4 0

3 years ago

Which type or types of motion is particle motion in surface waves?

kipiarov [429]

Answer:

circular motion

In surface waves, particles of the medium undergo a circular motion. They are neither longitudinal nor transverse, for in longitudinal and transverse waves, all the particles in the entire bulk of the medium move in a parallel and a perpendicular direction, respectively, relative to the direction of energy transport.

Explanation:

Hope this helps

4 0

3 years ago