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3 years ago

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A 10kg crate sits on a distant planet. The planet has the same mass as Earth, but 80% of the radius of Earth (Earth radius = 637

0km.) What is the approximate weight of the object on the distant planet?
A) 120N
B) 160N
C) 80N
D) 100N

1 answer:

Nataliya [291]3 years ago

8 0

The answer is d! Hope I helped you

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A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

snow_lady [41]

Answer:

9.81N

Explanation:

the force of attraction is given by

F=<u>GmM</u><u>/</u><u>R²</u><u> </u>

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

$\texttt{ }$

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

3 years ago

Read 2 more answers

10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by

rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

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3 years ago

Suppose a hydrogen atom in its ground state moves 130 cm through and perpendicular to a vertical magnetic field that has a magne

blagie [28]

Answer:a)1.11×10^-21Nm

b) 1.16×10^-3m

Explanation:see attachment

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4 years ago

The purpose of ________ experiment is to review the physical concepts and relationships associated with the flow of heat in and

shutvik [7]

Answer:

Calorimetry

Explanation:

In the calorimetry experiments, mix a known amount of a high temperature material, about 100 ° C, with an equivalent amount of water at room temperature. In a device thermally isolated from the environment, and the initial and final temperatures of liquid and solid are analyzed and energy conservation relationships, the specific heat of the unknown material can be found.

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3 years ago