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3 years ago

15

What is the kinetic energy of a jogger with a mass of 65.0 kg traveling at a speed of 2.5 m/s

2 answers:

Alexxandr [17]3 years ago

7 0

KE = 1/2mv^2

KE = 1/2(65)*2.5^2
KE = 32.5*2.5^2

KE = 203.125 Joules

The kinetic energy is equal to <span>203.125 J. </span>

Liono4ka [1.6K]3 years ago

5 0

This should be Physics, but anyways here you go!
Kinetiv Energy formula is:
K.E = $\frac{1}{2}$ mv²

Plug in the given.

K.E = $\frac{1}{2}$ (65)(2.5)²
= 203.125 J

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Which graph shows a mixed-degree system with exactly one solution?

polet [3.4K]

Answer : option D

we need to find out the graph shows a mixed-degree system with exactly one solution

The solution of two graphs will be the intersection of two graphs

In the first graph,

There is no intersection of both circles, So no solution

second graph,

The circle intersects the parabola at two point, so two solution for this graph

Third graph,

The circle and line does not intersects, so no solution

Fourth graph,

Both parabola intersects at one point, so one solution for this graph

So option D has exactly one solution

5 0

3 years ago

Charles bought 3 pounds of apples and a bag of chips that cost $2.60 He spent atonal of $6.35. Write and solve a equation to det

nasty-shy [4]

Answer:

$1.25 for a pound of apples.

Step-by-step explanation:

First I did, 6.35-2.60 to find the cost of JUST the apples. which was 3.75.

then I did 3.75/3 i divided them. It came out to 1.25. SO that means it costs $1.25 for a pound of apples.

8 0

3 years ago

What is the best first step in solving the equation 3 + = 5?

Dafna1 [17]

Answer:

5 - 3

Step-by-step explanation:

easy way to find out

5-3= your answer in one step

3 0

3 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

Describe how to find the number of weeks equal to 56 days

shutvik [7]

Umm im not sure maby do 7x diffrent numbers to find out idk sorry

7 0

4 years ago

Read 2 more answers