hlo anyone help
a stone is thrown upward with the kinetic energy of 10 joule if it goes up to a maximum height of 5 M find the initial velocity and mass of the stone??
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Answers

Answer 1

Explanation:

Here, kinetic energy of the body, K.E=10 J

Height attained by the body, h=5 m

h=5m

When the body attains maximum height, its kinetic energy is converted into potential energy.

P.E=K.E

⟹mgh=10

⟹m= 10/gh

= m = 10/(10×5) =0.2 kg (taking the value g as 10m/s²)

Now, ½mv²=10

0.2×v²=20

v²=20/0.2

v²=100

v=√100

v=10m/s.

hope this helps you.

Answer 2

[tex]\huge \bf༆ Answer ༄[/tex]

Let's solve ~

As we know the total energy [P.E + K.E = C] of the system remains constant throughout the motion,

So, When a stone was thrown initially it didn't had any potential energy (P.E = 0) but had kinetic Energy of 10 joules.

So, total energy = P.E + K.E = 0 + 10 = 10 joules

As per the given information, equate it with the formula.

[tex] \sf \dfrac{1}{2}m {v}^{2} = 10[/tex]

[tex] \sf m {v}^{2} = 20[/tex]

Now, As it approaches 5m height it comes to rest, Therefore velocity = 0. And since velocity = 0 then Kinetic Energy = 0

Now, let's find the Potential Energy at that point ~

[tex] \sf mgh[/tex]

[tex] \sf m \times 10 \times 5[/tex]

[tex] \sf50m[/tex]

And here, the Total energy = P.E + K.E = 10 Joules

So,

[tex] \sf50m + 0 = 10 [/tex]

[tex] \sf50m = 10[/tex]

[tex] \sf m = 10 \div 50[/tex]

[tex] \sf m = 0.2[/tex]

Therefore, mass of the object is 0.2 kg = 200 grams

Now, plug the value of mass (m) in the equation of kinetic Energy to find the initial velocity of the stone ~

[tex] \sf m {v}^{2} = 20[/tex]

[tex] \sf0.2 \times {v}^{2} = 20[/tex]

[tex] \sf {v}^{2} = 20 \div 0.2[/tex]

[tex] \sf v = \sqrt{100} [/tex]

[tex] \sf v = 10 \: \: ms {}^{ - 1} [/tex]

Hence, velocity of the particle at the beginning was 10 m/s


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A black hole forms as a star collapses when it ceases emitting radiation from fusion reactions. This occurs after the star has consumed hydrogen, helium, then through iron.

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468.42572 Wavelength In Metres

Explanation:

Answer:

[tex]\huge\boxed{\sf Wavelength= 470\ m}[/tex]

Explanation:

Given Data:

Speed of light = c = 3 × 10⁸ m/s (Constant)

Frequency = f = 640 kHz = 6.4 × 10² × 10³ Hz = 6.4 × 10⁵ Hz

Required:

Wavelength = λ = ?

Formula:

λ = c / f

Solution:

λ = 3 × 10⁸ / 6.4 × 10⁵

λ = 0.47 × 10³

λ = 4.7 × 10² m

λ = 470 m

[tex]\rule[225]{225}{2}[/tex]

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~AH1807

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Answer:

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Explanation:

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Answer:

[tex]\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}[/tex]

Explanation:

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The given parameters:

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The pail’s minimum speed at the top of the circle if no water is to spill out is; v = 4.722 m/s

We are given;

Mass of the pail of water; m = 2000 g = 2kg

Radius of the circle; r = 1 m

Tension exerted by string; T = 25 N

To calculate the net force on the pail of water at the top of the vertical circle, let us take equilibrium of forces to give;

W - T = ma

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W is weight = mg

T is tension on the string

Now, in circular motion, we know that;

Acceleration is; a = v²/r

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mg - T = m(v²/r)

Making v the subject of the formula gives us;

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A 30.4-newton force is used to slide a 40.0-newton crate a distance of 6.00 meters at constant speed along an incline to a vertical height of 3.00 meters. Calculate the work done by a friction force on the crate as it slides 6.00 m along the incline.

Answers

Hi there!

Since the crate is being slid at a constant speed, the forces sum to 0 N. In this instance, the following forces occur in the axis of interest:

Wsinθ = downward acceleration along incline due to gravity (N)

Fκ = kinetic friction force along incline (N)

A = applied force (N)

The acceleration due to gravity and friction force act in the same direction, so:

Wsinθ + Fκ = A

Solve for sinθ using right triangle trigonometry:

sinθ = O/H = 3/6 = 0.5

Rearrange the equation for the force of kinetic friction and solve:

Fκ = A - 0.5W

Fκ = 30.4 - 20 = 10.4 N

Now, recall that:

Work = Force × displacement (W = F × d)

Since the box's displacement is in the same axis as the force but OPPOSITE direction, we must use:

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Angle between displacement and friction force is 180°.

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Work done by friction = -Fd = -10.4(6) = -62.4 J

friction and normal force are examples of

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Answer:

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If a nonzero net force is acting on an object, then the object is definitely _____.

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Answers

An object is definitely being accelerated when a nonzero net force is acting on it: Option B.

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According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass.

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From one point on the ground, the
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mountain, the angle of elevation is
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the mountain. Find the height of the
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10.38°
一企一
14.67°
15,860 ft

Answers

Trigonometry allows to find the result for the question about the height of the mountain is:

The height mountain  is:  y = 9674.4 ft

Trigonometry allows finding relationships between the angles of a right triangle.

         [tex]tan \theta = \frac{y}{x}[/tex]  

Where θ is the angle, y the opposite leg (height) and x the adjacent leg (horizontal distance).

In the attachment we can see a diagram of the system. They indicate that for x  distance the angle is 14.67º  

         tan 14.67 = [tex]\frac{y}{x}[/tex]  

At the other point the angle is 10.38º.

        tan 10.38 = [tex]\frac{y}{x+15860}[/tex]

We look for the horizontal distance (x) with these equations.

        x tan 14.67 = (x + 15860) tan 10.38

        x tan 14.67 / tan 10.38 = x + 15860

        x 1,429 = x + 15860

        x 0.429 = 15860

        x = [tex]\frac{15680}{0.429}[/tex]

        x = 36955.3 ft

We calculate for the height.

        y = x tan 14.67

        y = 36955.3 tan 14.67

        y = 9674.4 ft

In conclusion using trigonometry we can find the result for the height of the mountain is:

The height is y = 9674.4 ft

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