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3 years ago

According to the Get Fit and Be active book, page 5 talks about the phases of exercise, what is step 2? Question 2 options:

Physics

1 answer:

never [62]3 years ago

8 0

Pretty sure it's warming up cuz first would stretch then warm up then exercise and then cool down

You might be interested in

Prove: <br>1) p=dhg <br>.........<br>

Lilit [14]

Explanation:

P = F/A

P = mg/A [ since F = mg ]

P = Vdg/A [ since m = Vd ]

P = Ahdg/A [ since V = Ah ]

P = hdg

6 0

3 years ago

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

Zigmanuir [339]

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$

7 0

3 years ago

Read 2 more answers

A heat engine extracts 42. 53 kj from the hot reservior and exhausts 17. 69 kj into the cold reservior. what is the work done?

ludmilkaskok [199]

The answer is 24.84kJ.

We apply the expression for the work done by the heat engine is,

$W=E_{i n}-E_{o u t}$ . Putting all given values in the equation we get the final answer.

What is heat engine?

A heat engine is a machine that uses heat to generate power. It draws heat from a reservoir, uses that heat to produce work, such as move a piston or lift weights, and then releases that heat energy into the sink.
We are given:The heat input is $E_{i n}=42.53 \mathrm{~kJ}$ . The heat output is $E_{o u t}=17.69 \mathrm{~kJ}$ .
The expression for the work done by the heat engine is, $W=E_{i n}-E_{o u t}$
Substituting the given values in the above expression, we will get $W =42.53 \mathrm{~kJ}-17.69 \mathrm{~kJ}$ =24.84kJ.
Thus, the work done by the heat engine is 24.84kJ.

To learn more about heat engine visit: brainly.com/question/15735984

#SPJ4

6 0

2 years ago

A ball is thrown vertically upward with a speed v. an identical second ball is thrown upward with a speed 2v. how many times hig

vladimir2022 [97]

It goes twice as fast as the first one. Twice

6 0

3 years ago

An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent

madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M

F = GmMx/[√(a² + x²)]³

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0

3 years ago