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4 years ago

An approved food thermometer reads from what

1 answer:

7 0

Temperature danger zone is between 41°F and 140°F. Keep hot food hot and coldfood cold. Always use a thermometer to check food temperatures. Potentially hazardous foods must pass through the temperature danger zone as quickly as possible.

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Which of the following examples best describe using an inclined plane

Alinara [238K]

The best possible Answer is c

8 0

3 years ago

Use the collision theory to explain how increasing the temperature of a reaction will affect the rate of the reaction.

kirill115 [55]

Increasing the temperature causes the particles in the reaction to become kinetically excited, hitting one another in increasing frequency. Increased collision among means faster rate or reaction.

7 0

3 years ago

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.0 vibrations

Goryan [66]

Answer:

0.47 m

Explanation:

$N$ = Number of vibrations = 37

$t$ = total time taken = 33 s

$T$ = time period of each vibration

frequency of vibration is given as

$f = \frac{N}{t} \\f = \frac{37}{33} \\f = 1.12$ Hz

$d$ = distance traveled along the rope = 421 cm = 4.21 m

$t$ = time taken to travel the distance = 8 s

$v$ = speed of the wave

Speed of the wave is given as

$v = \frac{d}{t}\\v = \frac{4.21}{8}\\v = 0.53 ms^{-1}$

$\lambda$ = wavelength of the harmonic wave

wavelength of the harmonic wave is given as

$\lambda = \frac{v}{f} \\\lambda = \frac{0.53}{1.12} \\\lambda = 0.47 m$

8 0

3 years ago

Read 2 more answers

A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun

Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100.0}{300}$

$t=0.33\ sec$

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s=0+\dfrac{1}{2}\times9.8\times(0.33)^2$

$s=0.53\ m$

Hence, The distance is 0.53 m.

5 0

4 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago