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svet-max [94.6K]
3 years ago
12

A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distan

ce d (at point A) and then released from rest, where the block undergoes harmonic motion. The floor is frictionless.The points labelled A and C are the turning points for the block, and point B is the equilibrium point.1)Which of these quantities are conserved for the spring and block system?mechanical energyx-direction linear momentumneither energy nor momentum
Physics
1 answer:
Burka [1]3 years ago
7 0

Answer:

Explanation:

When a spring attached with a block is stretched and released , the block starts moving under SHM. The elastic energy stored in it initially at the time of initial stretch is  repeatedly converted into kinetic energy and vice - versa while the body keeps moving under SHM. So we can tell that the mechanical energy of the system remains constant.

In the whole process the velocity of the body keeps changing in terms of both magnitude and direction . It happens because a variable force acts on the body constantly towards the equilibrium point  so its momentum also keeps changing all the time

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What are four types of pathogens and what type of disease or sickness do they cause?
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7 0
3 years ago
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

4 0
3 years ago
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
2 years ago
a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the
inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

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a = 9.8 m/s²

Find: t

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t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

Δx = 0.84 m

5 0
3 years ago
A
dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

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