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2 years ago

A 0.155 kg arrow is shot upward

Physics

2 answers:

solniwko [45]2 years ago

3 0

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

it can be derived using the expression below:

K.E = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Solve for K.E;

K.E = $\frac{1}{2}$ x 0.155 x 31.4 = 2.43J

katrin2010 [14]2 years ago

3 0

Answer:

KE = ½mv² = ½(0.155)(31.4)² = 76.4 J

Explanation:

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A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that

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Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels l as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height h it reached, we can calculate l using the sin(α) function:

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Replacing:

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2 years ago

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INTEGRATED SCIENCE-8th grade science Please help report cards are due for me at dec.18

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Our drinking water comes from lakes, rivers and groundwater. For most Americans, the water then flows from intake points to a treatment plant, a storage tank, and then to our houses through various pipe systems. A typical water treatment process.

Explanation:

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3 years ago

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marusya05 [52]

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

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=6×3.14×eta×0.05×25

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2 years ago

The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single

Serhud [2]

Answer:

The tension in the string = 2.065 N

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W = mg = (689/1000)9.8 = 6.75 N, D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

Thus the tension in the string = 2.065 N

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2 years ago