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3 years ago

If you attach a 100 g mass to the spring whose data are shown in the graph, what will be the period of its oscillations?

Physics

2 answers:

ycow [4]3 years ago

8 0

if for a force of 0.5 N we have a displacement of -0.02m we can calculate the elastic constant(k) with the formula F=-kx(F=0.5N x=-0.02)

k=F/x k=0.5/0.02=25N/m

now we can calculate the period by the formula

T=2π√(m/k)

in the mass we convert grams to kilograms so 100g=0.1kg

T=6.28√(0.1/25)⇒T=0.39seconds

aleksandr82 [10.1K]3 years ago

3 0

Answer:

try yahoo they are actually helpful at times

Explanation:

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3 years ago

Read 2 more answers

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

Semenov [28]

Answer:

A)Object 1 has the greater magnitude of its momentum.

B)The objects 2 have the greater kinetic energy.

Explanation:

For object 1 :

v₁ = v ,m₁ = 2 m

For object 2 :

$v_2=2\sqrt{v}$ ,m₂=m

We know that linear momentum given as

P = M V

M=Mass , V=Velocity

For object 1 :

P₁ =m₁ v₁

P₁ =2 m v

For object 2

$P_2=m_2v_2$

$P_2=2m\sqrt {v}$

We can say that object 1 have more momentum.

The kinetic energy

$KE_1=\dfrac{1}{2}m_1v_1^2$

$KE_1=\dfrac{1}{2}\times 2m\times v^2$

$KE_1=mv^2$

$KE_2=\dfrac{1}{2}m_2v_2$

$KE_2=\dfrac{1}{2}\times m\times 4v^2$

$KE_2=2mv^2$

Therefore both the object 2 have higher kinetic energy.

7 0

3 years ago

A body is projected upward making angle theta with horizontal .with the velocity of 300m/s .find its range and time of flight.

uysha [10]

Answer:

Answer is in the picture

8 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 23.0 and carries a cl

Bond [772]

Answer:

I = 33.04A

Explanation:

The magnetic field at the center of a loop of wire is proportional to the current flowing through the wire and inversely proportional to the radius

So therefore we have:

20/23 = I/38.0

We then solve for I

cross multiplying we have 20 x 38 = 23 x I

I = 20 x 38/ 23

I = 33.04A

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3 years ago

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball

Aleksandr [31]

Answer:

the velocity (magnitude and direction) of the ball) just before the collision is $v_i = 5.144 \ m/s$

The velocity (magnitude and direction) of the ball) just after the collision is $v_f = - 1.129 \ m/s$

Explanation:

According to the law of conservation of energy;

$m__{ball}} gh = \frac{1}{2}m__{ball}}v_i^2\\\\2* m__{ball}} gh = m__{ball}}v_i^2\\\\v_i^2 = \frac{2*m_{ball}gh}{m_{ball}}\\\\v_i^2 = 2gh\\\\v_i = \sqrt{2gh} \\\\v_i = \sqrt{2*9.8*1.35}\\\\$

$v_i = 5.144 \ m/s$

Thus; the velocity (magnitude and direction) of the ball) just before the collision is $v_i = 5.144 \ m/s$

Since, Air resistance is negligible, and the collision is elastic.

The equation for the conservation of momentum and energy can be expressed as:

$v_f = [\frac{m_1 -m_2}{m_1+m_2}]v_i\\\\v_f = [\frac{m_{ball} -m_{block}}{m_{ball}+m_{block}}]v_i\\\\v_f = [\frac{1.6 -2.5}{1.6+2.5}]*5.144\\\\$

$v_f = - 1.129 \ m/s$

The velocity (magnitude and direction) of the ball) just after the collision is $v_f = - 1.129 \ m/s$

8 0

3 years ago