PHYSICS COLLEGE

Look at the Kunoichi's of Naruto but as a Gang. Who do you think looks the baddest out of the group(but in a good way)?

My opinion is Hinata...just saying

Answers

Answer 1

Answer:

Answer:

hinata for sure

Explanation:

seems reasonable

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An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an instant, exceed 16 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit?A)8 sqrt 2 V

B) 16 sqrt 2 V

C) 256 V

D) 8

Answers

Answer:

A) $V_(rms)=8√(2) V$

Explanation:

Maximum voltage = $V_(max)=16 V$

Maximum voltage and rms voltage are related to each other by

$V_(max)=V_(rms) * √(2) \nV_(rms)=(V_(max))/( √(2))\nV_(rms)=(16)/(√(2)) \nV_(rms)=8√(2) V$

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 349 K, what is its radius at liftoff

Answers

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $(4)/(3) \pi r^(3)$

volume of balloon = $(4)/(3)$ x 3.142 x $24^(3)$ = 57913.34 m^3

using the gas equation,

$(P1V1)/(T1)$ = $(P2V2)/(T2)$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$(0.03*57913.34)/(200)$ = $(1*V2)/(349)$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $(4)/(3) \pi r^(3)$ = $(4)/(3)$ x 3.142 x $r^(3)$ = 3031.76

4.19 $r^(3)$ = 3031.76

$r^(3)$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TB
b) TA < TB
c) TA = TB
d) More information is needed

Answers

The final temperatures are such that TA > TB.

The specific heat capacity refers to the quantity of heat required to raise the temperature of 1 Kg of a body by 1K. The higher the specific heat capacity of a body, the higher the quantity of heat required to raise the temperature of the body and vice versa.

Hence, if the specific heat of substance A is greater than that of substance B and A and B are at the same initial temperature, when equal amounts of energy are added to them, the final temperature are such that TA > TB.

Learn more: brainly.com/question/1445383

Answer:

$m_A c_(pA) (T_(fA) -T) = m_B c_(pB) (T_(fB)- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_(pA) > c_(pB)$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_(iA)= T_(iB)$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_(pA) \Delta T_A = m_B c_(pB) \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_(pA) (T_(fA) -T_(iA)) = m_B c_(pB) (T_(fB)- T_(iB))$

And since $T_(iA)= T_(iB)= T$ we have this:

$m_A c_(pA) (T_(fA) -T) = m_B c_(pB) (T_(fB)- T)$

d) More information is needed

(The relation between the masses is not given)

Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?

Answers

Final answer:

Given the same initial linear speed, a solid ball, solid disk, and hoop will expend energy on both rotation and translation. The solid ball, having the lowest moment of inertia, uses the most energy for translation and, therefore, will travel the highest up an incline.

Explanation:

In the context of this problem related to physics, the deciding factor is the distribution of mass, which influences each object's moment of inertia. Objects set to roll tend to use energy in two ways: translation (moving along the incline) and rotation (spinning about their center). Moment of inertia essentially measures how much of the object's energy goes towards rotation.

For a solid ball, solid disk, and hoop with the same mass and radius, the hoop has the highest moment of inertia with all of its mass at the maximum distance from the center. Followed by the solid disk, with its mass spread evenly from the center to its edge. Lastly, the solid ball has the lowest moment of inertia as its mass is concentrated towards the center.

This means that, given the same initial linear speed, the hoop will expend most of its energy on rotation rather than moving up the incline (translation). The solid disk will have a more balanced distribution between translation and rotation, and finally, the solid ball will use the least amount of energy on rotation and the most on translation. As such, the solid ball will go the farthest up the incline.

Learn more about the Physics of Rotating Objects here:

brainly.com/question/11991361

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On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom?
b. Express your answer to two significant figures and include the appropriate units.
c. How far does the water move horizontally during this time?
d. Express your answer to two significant figures and include the appropriate units.

Answers

Answer :

a.3.29 m/s

b.3.3 m/s

c.14.8 m

d.15 m

Explanation:

We are given that

Initial Horizontal speed= $v_x=4.5 m/s$

Vertical component of initial speed= $v_y=0$

Vertical distance= $y=-53 m$

a. $s=ut+(1)/(2)gt^2$

Using the formula and g is negative therefore $g=-9.8m/s^2$

$-53=0-(1)/(2)(9.8)t^2$

$53=4.9t^2$

$t^2=(53)/(4.9) s$

$t=\sqrt{(53)/(4.9)}=3.29m/s$

b.Hundredth place is greater than 5 therefore, 1 will be added to tenth place and other digits on left side of tenth place remains same and digit on right side of tenth place replace by 0.

$t=3.3 m/s$

c.Horizontal acceleration= $a_x=0$

$x=v_xt=4.5* 3.29=14.8 m$

d.Tenth place 8 is greater than 5 therefore, 1 will be added to unit place and other digits on left side of unit place remains same and digit on right side of unit place replace by 0.

Horizontal distance=15 m

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?