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Umnica [9.8K]
3 years ago
13

A force of 100N stretches an elastic string to a total length of 20cm. if an additional force of 100N stretches the spring 5 cm

further, find the natural length of the spring if the elastic limit is not exceeded. Pls help ASAP​
Physics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

8cm

Explanation:

The general formula to find this is F/X where in this case the f is the total of the two forces (100+100) and x is the sum of the length ( 20+5) which together comes 200&25respectively then divide 200 with 25 which comes 8cm.

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Suppose there was 2 Newtons what would the mass of the cart be?
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2 pounds.................
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The specific gravity of an material is defined as the ratio of the density of the material to the density of water. The specific
Alexandra [31]

The percentage of the iceberg that is visible above the water's surface will be 11.5%.

<h3>What is density?</h3>

Density is defined as the mass per unit volume. It is an important parameter in order to understand the fluid and its properties. Its unit is kg/m³.

The mass and density relation is given as

mass = density × volume

Density of ice = 917 kg/m³

Density of the seawater=1025 kg/m³

The ratio by which the iceberg submerge is found as;

\rm  \frac{V_S}{V} =\frac{\rho}{\rho_0} \\\\ \frac{V_S}{V} =\frac{917}{1025} \\\\ \frac{V_S}{V} =.895

Hence,89.5 % of the ice is submerged.

The percentage of the iceberg that is visible above the water's surface is;

⇒1-.89.5 %

⇒11.55 %

Hence, the percentage of the iceberg that is visible above the water's surface will be 11.5%.

To learn more about the density refers to the link;

brainly.com/question/952755

5 0
3 years ago
A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.
g100num [7]

Answer:

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\

(A) the magnitude of its angular acceleration(rad/s^2);

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2

(B) Time of motion;

\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s

7 0
3 years ago
What is the distance time how can we find the speed of an object from its distance time graph​
CaHeK987 [17]

Answer:

speed is the gradient of the graph

7 0
3 years ago
Read 2 more answers
A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that a
Klio2033 [76]

Answer:

E_{square}=0.51v

i_{square}=2.98A

Explanation:

E_{mf}=-N*\frac{d\alpha}{dt}

E_{mf}=-N*A*\frac{d\beta}{dt}

N=1

E_{circle}=A_{circle}*\frac{d\beta}{dt}

E_{square}=E_{circle}*\frac{A_{square}}{A_{circle}}

E_{square}=E_{circle}(\frac{\frac{\pi^2}{4}*r^2}{\pi*r^2})

E_{square}=E_{circle}*\frac{\pi}{4}

E_{square}=0.65v*\frac{\pi}{4}

E_{square}=0.51v

i_{square}=i_{circle}*\frac{E_{square}}{E_{circle}}

i_{square}=3.8A*\frac{0.51v}{0.65v}

i_{square}=2.98A

8 0
3 years ago
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