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tia_tia [17]
2 years ago
8

A temperature of 34ºF is equal to ____________________ kelvin.

Physics
1 answer:
lara31 [8.8K]2 years ago
6 0
 Kelvin = (Fahrenheit - 32) * 5 / 9 + 273.15<span> 

then 
k = (34 - 32) *5/9 + 273.15
k = 274.26 </span>
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A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.
aleksley [76]
The acceleration that the cheeseburger experienced is 20 m/s^2.
6 0
2 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
When the speed of your car is doubled, by what factor does its kinetic energy increase?
Tju [1.3M]

Answer:

a. 4

Explanation:

Hi there!

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

Let´s see how would be the equation if the velocity is doubled (2 · v)

KE2 = 1/2 · m · (2 · v)²

Distributing the exponent:

KE2 = 1/2 · m · 2² · v²

KE2 = 1/2 · m · 4 · v²

KE2 = 4 (1/2 · m · v²)      

KE2 = 4KE

Doubling the velocity increased the kinetic energy by 4.

7 0
3 years ago
Which statement shows how to correctly convert from the mass of a compound in grams to the amount of that compound in moles?
expeople1 [14]

Answer:

mass (grams) x 1 =

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Explanation:

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3 years ago
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