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3 years ago

Kyle has a mass of 54 kg and is jogging at a velocity of 3 m/s.

Physics

1 answer:

Colt1911 [192]3 years ago

5 0

Answer:

Kinetic energy can be solved by using the following formula: Kinetic energy = (1/2)*m*v^2

Explanation:

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When is an atom stable? A. when it has a full outer orbit. B. when it has the same amount of elections as protons. C. when it ha

balu736 [363]

B when the electrons and protons are the same amount

3 0

3 years ago

Read 2 more answers

A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.

belka [17]

Answer:

Approximately $13\; {\rm N \cdot m^{-1}}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

Let $F_{\text{s}}$ denote the force that this spring exerts on the object. Let $x$ denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant $k$ of this spring would ensure that $F_\text{s} = -k\, x$ .

Note that the mass of the object attached to this spring is $m = 540\; {\rm g} = 0.540\; {\rm kg}$ . Thus, the weight of this object would be $m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}$ .

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, $F_{\text{s}} = 5.230\; {\rm N}$ .

The spring in this question was stretched downward from its equilibrium by:

$\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}$ .

(Note that $x$ is negative since this displacement points downwards.)

Rearrange Hooke's Law to find $k$ in terms of $F_{\text{s}}$ and $x$ :

$\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}$ .

3 0

2 years ago

A circular coil that has 100 turns and a radius of 10.0 cm lies in a magnetic field that has a magnitude of 0.0650 T directed pe

yulyashka [42]

Answer:

(a) 0.204 Weber

(b) 0.22 Volt

Explanation:

N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.

(a) Magnetic flux, Ф = N x B x A

Ф = 100 x 0.0650 x 3.14 x 0.1 0.1

Ф = 0.204 Weber

(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s

dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s

induced emf, e = N dФ/dt

e = N x A x dB/dt

e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V

6 0

3 years ago

· List the atomic numbers of the elements in Period 2.

PilotLPTM [1.2K]

Answer:

3,4,5,6,7,8,9,10

3. lithium

4. beryllium

5. boron

6. carbon

7. nitrogen

8. oxygen

9. fluorine

10. neon

those are the numbers in period 2 on the periodic table.

Explanation:

4 0

2 years ago

Read 2 more answers

You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 +9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km

7 0

3 years ago