Come on here its cool

A cook removes a one-gallon pot of hot soup from a stove and places it in an ice-water bath to cool. which is the best cooling p

Verizon [17]

I thank it is #2 or #1 hop this helps;)

5 0

3 years ago

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

tangare [24]

Answer:

Explanation:

Given

speed of Electron $u=2\times 10^7\ m/s$

final speed of Electron $v=4\times 10^7\ m/s$

distance traveled $d=1.2\ cm$

using equation of motion

$v^2-u^2=2as$

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

$(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}$

$a=5\times 10^{16}\ m/s^2$

acceleration is given by $a=\frac{qE}{m}$

where q=charge of electron

m=mass of electron

E=electric Field strength

$5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}$

$E=248.3\ kN/C$

5 0

3 years ago

A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s

balandron [24]

Answer: $233.23 m/s^2$

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

$1.5=0+\frac{gt^2}{2}$

t=0.55 s

Initial horizontal velocity at the time of break is given by u

$R=u\times t$

$10=u\times 0.55$

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

$a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2$

6 0

3 years ago

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

A)

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

B)

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

C)

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

2 years ago

A length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

ycow [4]

The equivalent resistance of resistors connected in parallel is
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$
In our problem, we have 5 identical resistor of resistance $R$ , so their equivalent resistance is
$\frac{1}{R_{eq}} = 5 (\frac{1}{R} )$
The problem also says that the equivalent resistance is $R_{eq}= 17 \Omega$ , so we can find the resistance R of each piece of wire:
$R=5 R_{eq} = 5 \cdot 17 \Omega = 85 \Omega$

In the initial wire, it's like the 5 pieces are connected in series, so the equivalent resistance of the wire is just the sum of the resistances of the 5 pieces:
$R_{wire} = 5 R = 5 \cdot 85 \Omega = 425 \Omega$

6 0

3 years ago