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4 years ago

A transfer of heat within a liquid or gas that involves warm particles moving in currents is

Physics

1 answer:

PolarNik [594]4 years ago

7 0

The answer is B. Convection

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THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!

poizon [28]

Answer:

391.67Hz

Explanation:

The fundamental frequency formula in string is expressed as;

Fo = V/2L

V is the velocity of the wave = 329m/s

L is the length of the string = 42cm = 0.42m

Substitute

Fo = 329/2(0.42)

Fo = 329/0.84

Fo = 391.67Hertz

Hence the fundamental frequency of a mandolin string is 391.67Hz

4 0

3 years ago

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

3 years ago

PLSS HELLPPP MEEE WILL MARK BRAINLIEST

Lesechka [4]

Answer:

a:s

b:s

c:r

d: r

e: s

f: s

g: s

Explanation:

8 0

3 years ago

In order to determine the velocity, you must know

Naddik [55]

You need to know the speed and direction of object

5 0

3 years ago

Read 2 more answers

(a) calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the radius of the earth is

kakasveta [241]

(a) The value of the gravitational acceleration is given by:

$g=\frac{GM}{r^2}$

where

$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

M is the Earth's mass

$r=6371 km=6.371 \cdot 10^6 m$ is the Earth's radius at the pole

If we use the value for g given by the problem, $g=9.830 m/s^2$ , and we rearrange the equation above, we find the value of the Earth's mass:

$M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg$

(b) The value we found for the Earth's mass is $5.982 \cdot 10^{24} kg$ , and we see that this value is slightly larger than the accepted value of $5.979 \cdot 10^{24} kg$ .

5 0

3 years ago