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k0ka [10]
3 years ago
12

An outside thermometer reads 57°F. What is this temperature in °C? Round your answer to the nearest whole number.

Physics
2 answers:
marishachu [46]3 years ago
3 0

Answer: It'd be 14.

Explanation:

The formula for this equation would be (57f-32)×5/9 which is equal to 13.889; and rounding that to the whole number would be 14.

Fiesta28 [93]3 years ago
3 0

Answer-

a) 14

Explanation:

Fahrenheit to Celsius Formula: (°F - 32) / 1.8 = °C

= [57-32]/1.8=C

=25/1.8=C

=13.8888888889

round off 13.8888888889=14

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A sample of an ideal gas is heated, and its kelvin temperature doubles. What happens to the average speed of the molecules in th
BabaBlast [244]

A sample of an ideal gas is heated, and its kelvin temperature doubles. The average speed of the molecules in the sample will increases by a factor of  \sqrt{2}

The root-mean square (RMS) velocity is the value of the square root of the sum of the squares of the stacking velocity values divided by the number of values. The RMS velocity is that of a wave through sub-surface layers of different interval velocities along a specific ray path.

Root mean square speed is a statistical measurement of speed.

The root mean square speed can be calculated as : V1 : \sqrt{3 R T / Mo}

if  temperature becomes double

let T1 is initial temperature

So ,  T2 = 2 * T1

now ,

Root mean square speed will be (V2) =  \sqrt{(3 R (2T)) / Mo}

                                                     = \sqrt{2} * \sqrt{3 R T / Mo}

                                                     = \sqrt{2} V1

Thus when temperature becomes double, the root mean square speed increases by a factor of  \sqrt{2}

To learn more about root mean square velocity here

brainly.com/question/13751940

#SPJ4

3 0
1 year ago
Match each type of wave to the way it moves.
oee [108]

Answer:

Transverse wave- Back and forth at right angles to the direction of the wave arrow.

longitudinal wave- bask and forth in the direction of the motion of the motion of the wave.

electromagnetic wave- two alternating waves moving at right angles to each other.

Explanation:

In a longitudinal wave, the particles vibrate at right angles in reference to the wave motion.

In a transverse wave, the particles vibrate parallel to the wave motion

Electromagnetic waves occur as a result of the interaction between two  waves and are normally transverse in nature.  

3 0
3 years ago
A 42 kg surfer jumps off the back of a 22 kg surfboard that is moving forward with a velocity of 5.2 m/s. After the surfer leave
Fittoniya [83]

Answer:

The surfer leave the surferboard with a velocity of 4.72[m/s]

Explanation:

This problem is related to the Conservation of Momentum, and it can be calculated using the following equation.

m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}^{'}+ m_{2}*v_{2}^{'}    \\

Where:

m1 = mass of the surfer = 42[kg]

v1 = velocity of the surfer before jumping = 5.2 [m/s]

m2 = mass of the surfboard = 22 [kg]

v2 = velocity of the surfboard before jumping = 5.2 [m/s]

Now after jumping

m1 = mass of the surfer = 42[kg]

v1' = velocity of the surfer after jumping = x

m2 = mass of the surfboard = 22 [kg]

v2' = velocity of the surfboard after jumping = 6.1 [m/s]

Now replacing in the equation.

(42*5.2)+(22*5.2)= (42*X)+(22*6.1)\\42*X = 198.6\\x = 4.72[m/s]

8 0
3 years ago
A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with
Lena [83]
REFER TO THE IMAGES for the SOLUTIONS TO YOUR PROBLEM. Each step will be explained here.

When you solve for velocities before or after collision, you need to remember the law of conservation of moment which can be expressed through this formula:

  BEFORE                   AFTER
m1v1+m2v2      =     m1v1 + m2v2

This basically means, the sum of momentum of 2 objects BEFORE collision is equal to the same 2 objects AFTER collision. 

The type of collision we have in your case is a 2D collision, where there is a gliding collision or they touch at an angle. So when you solve these type of problems, you have to consider the x and y components of motion. It makes things easier if you make a table like in the image to sort out your components.


STEP 1: COMPUTE FOR MOMENTUM BEFORE COLLISION for each OBJECT involved.
To solve for momentum, the formula is mass x velocity or mv:

STEP 1a: Body A: The problem states that before collision Body A is moving along the positive X-axis so the velocity will be +4.5 m/s. Notice that the velocity of the y component is 0 m/s. This is because BODY A is moving along the x-axis, with no mention that it deviated from it. 

STEP 1b: Body B: Body B is at rest before collision, that is why it is not moving at all, which means both x and y components are equal to 0.

STEP 1c: Get the sum of all X components and the Sum of Y components.

STEP 2:   COMPUTE FOR MOMENTUM AFTER COLLISION for each OBJECT involved.

Step 2a: BODY A: Notice that we now have an angle. hence the cos and sin. We do this because we are breaking or decomposing the diagonal velocity into its x and y component. To get the x-component you get the cos of the angle and multiply it to the momentum of the diagonal or overall velocity. For y-component, instead of cos, you get the sin. 

Step 2b: BODY B: Here we have unknowns, which we will derive later on. In this step, just plug in what you know into the formula. 

Step 2c: We already know the x and y momentum of the objects BEFORE collision and the law of conversation of momentum states that the momentum AFTER is the same. With this total we can move onto the next step.

STEP 3: Solving for the X and y component of the velocity of BODY B AFTER collision.

Step 3a: Using the formula given in the image, we plug in what we know first. We know the momentum of the BODY A already, so we can put it into the equation. We also know the sum of both momenta and we put that into the equation too. Now all we do is derive the formula. DO NOT FORGET THAT WE ARE TO USE ONLY X COMPONENTS. 

Step 3b: is the same as the previous step but instead, we use Y COMPONENTS only. 

STEP 4: Combining X and Y components to get the resultant velocity:
For this step you need to remember the Pythagorean theorem. This is applied here because when you draw a free body diagram of the velocities, it creates a right triangle where :
the hypotenuse represents the final velocity
the opposite side represents the y-component and;
the adjacent represents the x-component.

Refer to the image for the solution.

STEP 5: Solving for the angle at which BODY B is moving:
For this step you need to remember SOH CAH TOA to find the angle at which BODY B is moving. You already have all the components you need, including the hypotheses. You can use any of the functions, and they should come up with the same approximation. 

FINAL ANSWER: BODY B was moving at 1.35 m/s, 21 degrees above the x-axis. 

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4 years ago
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