Ask question

Ask question

All categories

1 year ago

13

Which is the higher temperature? (Assume temperatures to be exact numbers.)(a)1.)265°C or 265°F?2.)265°C3.)265°F 4.)They are the

same temperature..(b)1.)200°C or 370°F?2.))200°C3.)370°F 4.)They are the same temperature.

1 answer:

vazorg [7]1 year ago

7 0

Remember the following formulas to convert temperatures between Farenheit and Celsius:

$\begin{gathered} \text{ Farenheit to Celsius:} \\ C=\frac{5(F-32)}{9} \\ \\ \text{ Celsius to Farenheit:} \\ F=\frac{9}{5}C+32 \end{gathered}$

Convert all the temperatures to Celsius to compare them.

<h2>Part a)</h2>

Convert 265ºF to Celsius:

$\begin{gathered} C=\frac{5(F-32)}{9} \\ \\ =\frac{5(265-32)}{9} \\ \\ =129.444... \end{gathered}$

Then, 265ºF is equal to 129.444...ºC. Then, 265ºC is higher than 265ºF.

The correct choice is option 2) 265ºC.

Part b)

Convert 370ºF to Celsius:

$\begin{gathered} C=\frac{5(F-32)}{9} \\ \\ =\frac{5(370-32)}{9} \\ \\ =187.777...ºC \end{gathered}$

Then, 370ºF is equal to 187.777...ºC. Then, 200ºC is higher than 370ºF.

The correct choice is option 2) 200ºC.

You might be interested in

Calculate the momentum for the following scenario: a bowling ball is moving at a velocity of 15 m/s and has a mass of 8.00 kg. *

svetlana [45]

Answer:

120 kg • m/s

Explanation:

<em>p = mv</em>

15 × 8.00 = 120

<em />

5 0

3 years ago

How does Newton's second law of motion can be used to calculate the acceleration of an object?

Alex777 [14]

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

4 0

4 years ago

A 4 kg billiard ball moving on a horizontal surface has a speed of 16 m/s when it strikes a horizontal coiled spring is brought

ruslelena [56]

Answer:

spring constant of the spring is 1820.44 N/m

Explanation:

given data

ball mass = 4 kg

speed = 16 m/s

distance = 0.75 m

to find out

spring constant of the spring

solution

we know that kinetic energy of ball = energy store in spring as compression

so we can express it as

0.5 × m × v² = 0.5 × k × x² ....................1

so put here value we get spring constant k

m × v² = k × x²

4 × 16² = k × 0.75²

solve it we get

k = 1820.44 N/m

so spring constant of the spring is 1820.44 N/m

6 0

4 years ago

Read 2 more answers

A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the

V125BC [204]

Answer:

91.48N/m

Explanation:

In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;

$\frac{1}{f}$ = $2\pi\sqrt{\frac{m}{k} }$ -----------(i)

From the question;

f = 3.26 Hz

m = 0.218kg

Substitute these values into equation (i) as follows;

$\frac{1}{3.26}$ = $2\pi\sqrt{\frac{0.218}{k} }$ [<em>Square both sides</em>]

( $\frac{1}{3.26}$ )² = ( $2\pi$ )²( $\frac{0.218}{k}$ )

( $\frac{1}{10.6276}$ ) = $4\pi$ ²( $\frac{0.218}{k}$ ) [<em>Take </em> $\pi$ <em> to be 3.142</em>]

( $\frac{1}{10.6276}$ ) = $4(3.142)$ ²( $\frac{0.218}{k}$ )

( $\frac{1}{10.6276}$ ) = $39.488$ ( $\frac{0.218}{k}$ )

( $\frac{1}{10.6276}$ ) = ( $\frac{8.608}{k}$ ) [<em>Switch sides</em>]

( $\frac{8.608}{k}$ ) = ( $\frac{1}{10.6276}$ ) [<em>Re-arrange</em>]

( $\frac{k}{8.608}$ ) = ( $\frac{10.6276}{1}$ ) [<em>Cross-multiply</em>]

k = 8.608 x 10.6276

k = 91.48N/m

Therefore, the spring constant of the spring is 91.48N/m

4 0

4 years ago

A horizontal pipe narrows from a radius of 0.250 m to 0.1000 m. If the speed of the water in the pipe is 1.00 m/s in the larger-

umka21 [38]

Answer:

6.25 m/s

196 kg/s

Explanation:

The areas of the pipe from big to small:

$A = \pi R^2 = \pi*0.25^2 = 0.196 m^2$

$a = \pi r^2 = \pi*0.1^2=0.0314m^2$

As the product of speed and cross-section area is constant, the speed in the smaller pipe would be

$AV = av$

$v = \frac{AV}{a} = \frac{0.196 * 1}{0.0314} = 6.25 m/s$

The mass flow rate would be:

$\dot{m} = \ro AV = 1000 * 0.196 * 1 = 196 kg/s$

4 0

3 years ago