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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The Law of Exponents applies here

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3^15 x 3^3 = 3^18

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Anyone please.??????​

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Answer:

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 = ?

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λ = c / f

Solution:

λ = 3 × 10⁸ / 6.4 × 10⁵

λ = 0.47 × 10³

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

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~AH1807

From one point on the ground, the
angle of elevation of the peak of a
mountain is 10.38°, and from a
point 15,860 ft closer to the
mountain, the angle of elevation is
14.67º. Both points are due south of
the mountain. Find the height of the
mountain.
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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Answer:

The distance of one lap on this track from the start line to the finish line is 400 meters. Two laps around the track is 800 meters, half a lap is 200 meters, and so on.

    Remember, displacement is the direction from the starting point and the length of a straight line from the starting point to the ending point. If a runner was going to run 400 meters (one lap around the track), they would start and stop at the same place on the track. Therefore, their displacement would be 0.

 

Explanation:

hope this helped

Answer:

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

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has no displacement. If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has 50 meters of displacement. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, Gabi had the greater displacement.

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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)

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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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We have the final answer as

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Answers

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We are given;

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

Radius of the circle; r = 1 m

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