List some activities that influence health-related fitness

How much work does the electric field do in moving a proton from a point with a potential of +130 V

In-s [12.5K]

Explanation:

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6 0

3 years ago

Which of the following is Not true about the noble gases

Alexandra [31]

Noble gases are not highly reactive

3 0

4 years ago

Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li

tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

W=9.81x Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: $(F_{J} +F_{T} =9.81x n)$

$F_{T} (510)=9.81x$ (210+21.5)

$F_{T} (510)=9.81 x (231.5)$

$F_{T} =4.452x N.$

$F_{J} =9.81x-4.452x$

$F_{J} =5.358 xN$

Therefore $F_{J} \geq F_{T}$

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0

2 years ago

You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you

Nataliya [291]

Answer:

Please check the attached file for the diagram

Explanation:

The velocity of the of the rowboat $V_{tot}$ through the river is the resultant velocity. It is obtained taking a vector sum of the velocity in still water and the velocity of the river.

There are several ways to take this vector sum, but the question makes it simple for us to use Pythagoras's theorem because the East and North directions are perpendicular to each other.

Hence;

$V_{tot}^2=V_{still}^2+V_{w}^2\\V_{tot}^2=3^2+4^2$

$V_{tot}=\sqrt{3^2+4^2}\\ V_{tot}=\sqrt{25}=5m/s$

6 0

3 years ago

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl

Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0

3 years ago