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3 years ago

7

Simplify 6.25 − 8.please help

1 answer:

tatyana61 [14]3 years ago

7 0

6.25 - 8 = -1.75

Hope this helps

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The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it

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Answer:

The total Mechanical energy will be zero

Explanation: Escape velocity is the velocity required by a free object in order to overcome the impact of the force of gravity. The total mechanical energy of an object is the total energy possessed by an object which includes its kinectic and potential energy.

since the object is moving at an escape velocity which is 11.2m/s the object will be assumed to be weightless

Etotal = kinetic energy + potential energy

kinetic energy= 1/2*M*V*V

Potential energy=MGH

Etotal=1/2*0*11.2*11.2+0*0*0

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3 years ago

A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th

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Answer:

$L^2(\dfrac{8m}{3}+m_r)$

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

$m_r$ = Mass of ring

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$\dfrac{mL^2}{3}$

For 8 spokes

$8\dfrac{mL^2}{3}$

Moment of inertia of ring

$m_rL^2$

Total moment of inertia

$8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)$

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is $L^2(\dfrac{8m}{3}+m_r)$ .

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2 years ago

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

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2 years ago