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3 years ago

If a 4N weight is hung on a spring, and it extends by 0.2m, what is the spring constant (k)?

Physics

2 answers:

tresset_1 [31]3 years ago

6 0

Answer:

20 N/m

Explanation:

4/0.2 = 20 N/m

/ = divide

Naddika [18.5K]3 years ago

4 0

Answer: 200 N/m

Explanation:

The Gravitational spring energy(Us) is equal to 1/2kx^2. So we have x as .2 m and Us as 4 N. So 4 N = 1/2 * k * .2^2. So now we solve for K and get 200 N/m.

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tickets at a museum cost 17 dollars each for a field trip, the museum offers a 4 doller discount on each ticket. How much will t

Marizza181 [45]

Answer:

$416 sounds like the best answer.

Explanation:

The tickets started off at $17, but a $4 discount for EACH ticket bought.

so 17-4 is 13

now tickets cost $13 each.

Multiply it by 32.

13x32=416

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3 years ago

Just the answer (PLSS)

Salsk061 [2.6K]

The miracle year for Albert Einstein was the year 1905 within which he published so many renowned papers.

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2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid