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vova2212 [387]
4 years ago
14

Why does the equilibrium position of the spring change when a mass is added to the spring? Will the mass oscillate around the ne

w equilibrium position of the spring or the previous position without a mass attached to the spring? If the equilibrium position of the spring changes by 20 cm (assuming no initial mass) when a mass is added to the spring with constant 4.9 kg/s^2, what is the mass of the object attached to the spring?
Physics
1 answer:
abruzzese [7]4 years ago
7 0
-it is because force is acted against gravity by the spring when we keep the mass on it
-it depends upon the spring constant
-i will answer it later!!
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A hot-air balloon is rising straight up with a speed of 1.30 m/s. A ballast bag is released from rest relative to the balloon wh
DiKsa [7]

Answer:

The time before the ballast hits the ground is 1.186918s

Explanation:

We assume that the upward direction is positive.

The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.

The balloon has a velocity (V0y) = 3m/s

When the bag is released, it has a height of 5.36m

To touch the ground it makes a displacement of y = - 5.36m

The motion of the bag can be described as followed:

y = V0t + 1/2 * at²

t² + (2V0/a)*t - (2y/a) = 0

We can solve this equation for t:

t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2

t = -(V0/a) ± √((V0/a)² + (2y/a))

In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)

t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))

t = 0.132653 ± 1.054266

t= 1.186918

The time before the ballast hits the ground is 1.186918s

6 0
3 years ago
Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can
s344n2d4d5 [400]
The answer is a in the center of active volcanos
8 0
3 years ago
Read 2 more answers
Given the displacement vector D = (2î − 6ĵ) m, find the displacement vector (in m) R so that D + R = −5Dĵ.
shusha [124]

Answer:

R = (-2î − 6ĵ + 10k^)m

Explanation:

We are given;

D = (2î − 6ĵ)

Now, we want to find R such that,

D + R = −5Dĵ

Plugging in (2î − 6ĵ) for D in the R equation gives;

(2î − 6ĵ) + R = -5(2î − 6ĵ)j

(2î − 6ĵ) + R = -10k + 0

This is because in vector multiplication, i × j = k and j × j = 0

Thus;

(2î − 6ĵ) + R = -10k

Making R the subject gives;

R = -2î − 6ĵ + 10k^

Thus, the displacement vector R is;

R = (-2î − 6ĵ + 10k^)m

7 0
4 years ago
Which two types of potentional energies would be types of mechancial energies
EastWind [94]

Answer:

Gravitation Potential Energy and Elastic Potential Energy

Explanation:

PE=abbreviation for potential energy

PE is a mechanical energy like KE is. There are two types of PE in high school physics (do not talk about other PE's like chemical PE).

Elastic PE is like the stored energy for example in a pole vault competition, the Elastic PE is stored in the pole before the vaulter goes over the desired height. (Energy stored in ELASTIC object ONLY) (commonly on the 11th grade regents springs are used to represent elastic PE)

Gravitational PE can be calculated through the formula PE=mgh although some people substitute the PE with a U. (ENERGY STORED IN OBJECT'S HEIGHT)

6 0
3 years ago
The following charges are located inside a submarine: 2.75 μC, −9.00 μC, 27.0 μC, and −43.5 μC.(a) Calculate the net electric fl
Kaylis [27]

Answer with Explanation:

We know that by Gauss's Law

E\cdot Flux=\frac{q_{inside}}{\epsilon _o}

Where

\epsilon _o is permitivity of free space

q_{inside} is the net charge in the hull

Part a)

Applying the given values we get

E\cdot Flux=\frac{(2.75+27-9-43.5)\times 10^{-6}C}{8.85\times 10^{-12}}\\\\\therefore E\cdot flux=-2.5\times 10^{6}

Part b)

Since the sign of the electric flux is negative we conclude that the electric field lines entering the submarine are more than the electric field lines leaving the submarine. Since negative sign implies that flux by negative charges is dominant.

7 0
3 years ago
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