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1 year ago

5

The temperature in degrees Celsius (C) can be converted to degrees Fahrenheit (F) using the formula F(C)= 9 5 C k, where k is a

constant. Find k, if to C=â’40oC corresponds F=â’40oF.

1 answer:

Lyrx [107]1 year ago

5 0

The value of the constant K in the temperature equation is 32.

The given parameters:

$F (C) = \frac{9}{5} C \ + K$
40⁰C corresponds to 104 ⁰F

The value of K for which the value of the temperature at 40 degrees Celsius corresponds to 104 degrees Fahrenheit is calculated as follows;

$F(C) = \frac{9}{5} C \ + \ K \\\\104 = \frac{9}{5} (40) \ + \ K\\\\104 = 72 \ + \ K\\\\104 - 72 = K\\\\32 = K$

Thus, the value of the constant K in the temperature equation is 32.

Learn more about temperature conversion here: brainly.com/question/17408526

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When driving a Toyota avensis car along a straight road for 16.5km at

Vitek1552 [10]

The average velocity of the whole journey will be total distance covered divided by the total time. It will be approximately equal to 8 m/s. The right answer is option B

<h2>VELOCITY</h2>

Velocity is the distance travelled in a specific direction. While the average velocity of the whole journey will be total distance covered divided by the total time

When driving a Toyota avensis car along a straight road for 16.5km at

50km/h,

The velocity = 50 km/h

Distance = 16.5 km

Use the speed formula to calculate time.

Speed = distance / time

Time = distance / speed

Time = 16.5 / 50

Time = 0.33 s

If over the next 20min, you walked another 2.5km further along the road for a petrol station, Then,

average velocity = Total distance covered divided by total time taken.

Where

The time t = 20/60 = 0.333 h

Total time = 0.33 + 0.3333

Total time = 0.6633333

Total distance = 16.5 + 2.5

Total distance = 19 km

Average velocity = 19 / 0.66333

Average Velocity = 28.64 km/h

Now convert Km/h to m/s

(28.6432 x 1000) / 3600

286432 / 3600

7.956m/s

Therefore, the average velocity of the whole journey from beginning of the drive to the arrival at the filling station will be approximately 8 m/s

Learn more about velocity here: brainly.com/question/6504879

7 0

2 years ago

A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

postnew [5]

Answer:

r=15.53 nm

$F=9.57\times 10^{-13}N$

Explanation:

Lets take electron is in between iron and uranium

Charge on electron $q_1= -1.602\times 10^{-19}C$

Charge on iron $q_2= 2\times 1.602\times 10^{-19}C$

Charge on uranium $q_3= 1.602\times 10^{-19}C$

We know that force between two charge

$F=K\dfrac{q_1 q_2}{r^2}$

$K=9\times 10^9\dfrac{N-m^2}{c^2}$

For equilibrium force between electron and iron should be force between electron and uranium

Lets take distance between electron and uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

$K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}$

$K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}$

$q_2= 2\timesq_1,q_3=q_1$

$\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}$

So r=15.53 nm

So force

$F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}$

$F=9.57\times 10^{-13}N$

7 0

2 years ago

A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.

belka [17]

Answer:

Approximately $13\; {\rm N \cdot m^{-1}}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

Let $F_{\text{s}}$ denote the force that this spring exerts on the object. Let $x$ denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant $k$ of this spring would ensure that $F_\text{s} = -k\, x$ .

Note that the mass of the object attached to this spring is $m = 540\; {\rm g} = 0.540\; {\rm kg}$ . Thus, the weight of this object would be $m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}$ .

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, $F_{\text{s}} = 5.230\; {\rm N}$ .

The spring in this question was stretched downward from its equilibrium by:

$\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}$ .

(Note that $x$ is negative since this displacement points downwards.)

Rearrange Hooke's Law to find $k$ in terms of $F_{\text{s}}$ and $x$ :

$\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}$ .

3 0

2 years ago

A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about

Stolb23 [73]

Answer:

Explanation:

The bead is moving on a vertical circular path so it must have a centripetal force towards the centre.

This force is equal to m v² / r

v is velocity of bead and r is radius of the circular path.

The vertical hoop is also rotating about a vertical axis passing through the centre at frequency f so the bead will experience a cetrifugal force due to rotation of the hoop. Its value is

m ω² r . Only at the point o degree and 180 degree , these forces are opposite to each other so at these points , the bead will be in equilibrium .

mv² / r = m ω² r

v² = ω² r²

v = ω r

= 2π f r

= 2 x 3.14 x 2 x 0.22

v = 2.76 m /s

For the bead to rise upto point θ = 90 degree , height achieved is radius R

required velocity = √ 2gR

= √ 2x 9.8x.22

= 2.076 m/s

This velocity is less than the velocity calculated earlier so the bead will be able to ride the required height.

v = 2.76 m/s

5 0

2 years ago

Gamma ray technology can be used to do which of the following?

Rudiy27

Answer:

b

Explanation:

radiation can treat tumors.

5 0

2 years ago