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Tanzania [10]
3 years ago
7

Which shows the formula for converting from degrees Celsius to degrees Fahrenheit?

Physics
2 answers:
lora16 [44]3 years ago
6 0

Explanation :

Degree Celsius and degree Fahrenheit are two units of measuring temperature.

The correct formula used to convert degrees Celsius to degrees Fahrenheit is :

^{\circ}F=\frac{9}{5}\times\ ^{\circ}C+32

Where,

^0F is temperature in Fahrenheit

^0C is temperature in Celsius

Hence, the correct option is (a).        

LenKa [72]3 years ago
4 0
Its simple, you have to plot in an example, but dont get confused with K, the answer its b

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4 0
3 years ago
an 80 kg cart goes around the inside of a vertical loop of a roller coaster. the radius of the loop is 5m and the cart moves at
GarryVolchara [31]

Answer:

F=  224 N

Explanation:

Given that

mass ,m = 80 kg

Radius ,r= 5 m

speed at the top v= 8 m/s

The force at the top = F

Now by using the Second law of Newton's

F+mg=\dfrac{mv^2}{r}

Now by putting the values

F+80\times 10 = \dfrac{80\times 8^}{5}

Take  g = 10 m/s²

F= \dfrac{80\times 8^2}{5}-800

F=  224 N

Therefore the force exerted by the track at the top position will be 224 N.

3 0
3 years ago
Two balls have masses 18 kg and 47 kg. The 18 kg ball has an initial velocity of 76 m/s (to the right) along a line joining the
BigorU [14]

Answer:

The final velocity of 18 kg ball is V_{2} = 42.09 \frac{m}{sec}

Explanation:

Mass of first ball m_{1} = 18 kg

Mass of second ball m_{2} = 47 kg

Initial velocity of 18 kg ball V_{1} = 76 \frac{m}{sec}

Initial velocity of 47 kg ball = 0

Final velocity of 18 kg ball V_{2} = ??

Final velocity of 18 kg ball  is given by the formula

V_{2} = \frac{2 m_{1} V_{1} }{m_{1} + m_{2}  }

Put all the values in above formula we get

V_{2} = 2 × 18 × \frac{76}{65}

V_{2} = 42.09 \frac{m}{sec}

Thus, the final velocity of 18 kg ball is V_{2} = 42.09 \frac{m}{sec}

3 0
2 years ago
A child is playing with a spring toy, first stretching and then compressing it.
Reika [66]
Sorry, I’m only in 6th Grade, I don’t know the answer to this question.
5 0
3 years ago
A meter stick swinging from one end oscillates with a frequency f0. What would be the frequency, in terms of f0 , if the bottom
snow_lady [41]

To solve this problem we will proceed to define the Period of a stick, then we will define the frequency, which is the inverse of the period. We will compare the change suffered by the new length and replace that value. The Time period of meter stick is

T = 2\pi\sqrt{ \frac{L}{g}}

Here,

L = Length

g = Gravity

At the same time the frequency is

f = \frac{1}{T}

Therefore the frequency in Terms of the Period is

f_0 = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

If bottom third were cut off then the new length is

L' = \frac{2}{3} L

Replacing this value at the new frequency we have that,

f ' = \frac{1}{2\pi} \sqrt{\frac{3g}{2L}}

f' = \sqrt{\frac{3}{2}}(\frac{1}{2\pi}\sqrt{\frac{g}{L}})

Finally,

\therefore f' = \sqrt{\frac{3}{2}}f_0

3 0
3 years ago
Read 2 more answers
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