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lubasha [3.4K]
3 years ago
9

The girl is using her arm to lift a weight. Her arm is acting as a _______.

Physics
2 answers:
Maurinko [17]3 years ago
8 0
Force.

Pulling and pushing are also ways of exerting force on an object.
devlian [24]3 years ago
6 0
Her arm acts as a pulley
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In which of the following cases would you feel weightless?
Dafna11 [192]
Falling from an airplane.
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3 years ago
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How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
Who was the discoverer of the neutron
Pavel [41]

Answer:

James Chadwick

Explanation:

In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron

6 0
2 years ago
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You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type
alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

\tau = F d sin \theta

where

F is the magnitude of the force

d is the length of the arm

\theta is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

\theta=90^{\circ}

Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

7 0
3 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
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