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3 years ago

If the mass of your father is 70kg, what is his weight (N)?

Mathematics

1 answer:

Airida [17]3 years ago

3 0

Answer: 686 N

Step-by-step explanation:

Hi!

Second Newton's law is: F=m*a, where F is force, m is mass, and a acceleration

On the Earth's surface, weight is the gravity force W=m*g, where g=9.8 m/s² is the acceleretion of gravity on Earth. So the weight of someone with mass m=70 kg is W=70*9.8 kg*m/s² = 686 N.

The unit N (Newton) is defined as 1 N = 1 kg*m/s²

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The height of a box cereal is about how many inches. 3in, 7in, 15in, or 27in

Tomtit [17]

It would be 15 In Because It Is A Little Longer Than A Ruler (foot) But Not Quite 27 Inches.

8 0

3 years ago

Read 2 more answers

Find the distance of a plane that travels at an average speed of 200mi/hr<br> for 8.5 hours

asambeis [7]

Answer:

1800 mi

Step-by-step explanation:

Recall that (distance) = (rate)(time).

Here, (distance) = (200 mi/hr)(8.5 hr) = 1800 mi

6 0

3 years ago

Question 2 (Essay Worth 10 points)

sashaice [31]

Ryan is correct, the mean is about 8.1.

To find the mean, you have to add up all the values, then divide by the total amount.

For Sample A:
60 x 6 + 90 x 7 + 145 x 8 + 150 x 9 + 55 x 10 = 4050

If you add up all the adults, you have 500.

Dividing 4050 by 500 = 8.1

Sample B will produce a very similar result.

5 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

−5y=−5<br> 7x+6y=7<br> <br> Is (5,1) a solution of the system

enot [183]

Answer:no, (5,1) is not a solution for the equation. (1/7,1) will be a solution for the equation.

Step-by-step explanation:

one way to do it is to plug 1 as the y-value into the first equation which does work but when you plug 5 as the x-value and 1 as the y-value in the second equation, it will get you to 41 which does not match so it will not be an equation. the other way to check is to solve by substitution which for the first equation, you divide both side by -5 and get y=1 then substitute y with 1 in the second equation and subtract both side by 6 and get 7x=1 and divide both side by 7 to get 1/7. the y-value match but the x-value don't so (5,1) is not a solution.

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3 years ago