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A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500 - kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency ω, the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (e) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.

Answer:

(A) 0.54 kg

(B) 15.5 m/s west

Explanation:

Mass is conserved.

M = m₁ + m₂

0.86 kg = 0.32 kg + m₂

m₂ = 0.54 kg

Momentum is conserved (take east to be positive).

Mv = m₁v₁ + m₂v₂

(0.86 kg)(-6 m/s) = (0.32 kg)(10 m/s) + (0.54 kg) v₂

v₂ = -15.5 m/s

Answer:

the velocity of the big fish after the launch is 6 m/s.

Explanation:

Given;

mass of the big fish, m₁ = 5 kg

velocity of the big fish, u₁ = 8 m/s

mass of the small fish, m₂ = 1 kg

velocity of the small fish, u₂ = -4 m/s

Let the final velocity of the big fish after launch = v

Apply the principle of conservation linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

5 x 8   + 1 x (-4) = v(5 + 1)

40 - 4 = 6v

36 = 6v

v = 36/6

v =  6 m/s.

Therefore, the velocity of the big fish after the launch is 6 m/s.

Answer:

Air

Explanation:

The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.

Answer:

[tex]1.714\ \text{A}[/tex]

Explanation:

F = Magnetic force = 0.018 N

B = Magnetic field = 0.03 T

L = Length of wire = 35 cm

[tex]\theta[/tex]  = Angle between current and magnetic field = [tex]90^{\circ}[/tex]

Magnetic force is given by

[tex]F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}[/tex]

The magnitude of the current is [tex]1.714\ \text{A}[/tex].

Answer:

the velocity of the swimmer at the bottom of the slide is 6.26 m/s

Explanation:

The computation of the velocity of the swimmer at the bottom of the slide is given below:

v = √2gh

= √2 × 9.8 × 2

= 6.26 m/s

Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s

The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change  with reference to time is referred to as the velocity

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

The velocity is found as;

[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]

Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.

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poste en français s’il vous plaît

Answer:

[tex]4.46\times 10^{-4}\ \text{V}[/tex]

Explanation:

B = Magnetic field = [tex]5\times 10^{-5}\ \text{T}[/tex]

r = Radius of rim = 16 m

t = Time = 90 seconds

A = Area of rim = [tex]\pi r^2[/tex]

EMF is given by

[tex]\varepsilon=\dfrac{BA}{t}\\\Rightarrow \varepsilon=\dfrac{5\times 10^{-5}\times \pi\times 16^2}{90}\\\Rightarrow \varepsilon=0.000446=4.46\times 10^{-4}\ \text{V}[/tex]

The magnitude of the induced emf between the hub and the rim is [tex]4.46\times 10^{-4}\ \text{V}[/tex].

Answer:

[tex]\tau=0.03\ N-m[/tex]

Explanation:

Given that,

Force acting, F = 6N

The radius of the path, [tex]r=10^{-2}\ m[/tex]

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find the amount of torque acting on the object. The formula for torque is given by :

[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]

So, the required torque is equal to 0.03 N-m.

I believe you are incorrect. A weathered mountain would appear more jagged.

I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.

If this is incorrect, please, don't refrain to tell me.

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Explanation:

Answer:

in this case the weight of the vehicle does not change , consequently the friction force should not change

Explanation:

The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation

          fr = μ N

          W-N= 0

          N = W

as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change

Answer:

The mass of the rod is 16 kg.

Explanation:

Given that,

The length of a rod, L = 3 m

The moment of inertia of the rod, I = 12 kg-m²

We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :

[tex]I=\dfrac{ML^2}{12}[/tex]

Where

M is mass of the rod

[tex]M=\dfrac{12I}{L^2}\\\\M=\dfrac{12\times 12}{(3)^2}\\\\M=16\ kg[/tex]

So, the mass of the rod is 16 kg.

Answer:

High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.

Answer:

As we can see, a string is attached with block A, and three string is folded with ply which is attached with B

x  

B

​  

=3x  

A

​  

Now differentiate with respect to x

V  

B

​  

=3V  

A

​  

Given,

V  

A

​  

=0.6m/s(totheright)

So,

V  

B

​  

=0.6×3

=1.8m/s(downward)

​

Explanation:

IF THE ANSWER IS RIGHT PLZ GIVE ME BRAINLIEST

THANK U

HAVE FUN AND BE SAFE

Answer:

Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.

Explanation:

Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose

Answer:

Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.

Answer:

Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.

The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.

In old times, grain had to be beaten out of the grass on a Threshing Floor.

Answer:

B: Amplitude

Explanation:

When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.

The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.

Answer:

The angle

Explanation:

Answer:

Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.

Answer:

See explanation

Explanation:

Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.

As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.

Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

       [tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]  

       [tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]

       [tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]

       [tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]

       [tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]

       [tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]

       [tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]

       [tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]

       [tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]

       [tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]

        [tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]

Answer:

   ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]    

Explanation:

This is an interesting problem, no data is given, so the result is a general expression.

Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂

we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved

initial instant. Just before the crash

           L₀ = L₁ + L₂

with

           L₁ =  I₁ w₁

the moment of inertia of a disc with an axis passing through its center is

           I₁ = ½ m₁ r₁²

we substitute

           I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂

final instant. Right after the crash

         L_f = I w

in angular momentum it is a scalar quantity, so it is additive

         I = I₁ + I₂

angular momentum is conserved

         L₀ = L_f

         I₁ w₁ + I₂ w₂ = I w

         w =  [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex]            (1)

We already have the angular velocities of the system, let's find the kinetic energy of it

initial

          K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²

final

          K_f = K = ½ I w²

the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses

         ΔK = K_f -K₀

         ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²)           (2)

In this chaos we know all the values ​​for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated

Now if all this variation of energy turns into heat

         Q = ΔK

         m_{total} c_e ΔT = ΔK

where the specific heat of the bear discs must be known, suppose they are of the same material

         ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]               (3)

to make a special case, we suppose some data

the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC

we look for the angular velocity

          I₁ = I₂ = I₀

          I = 2 I₀

we substitute in 1

           w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io

           w = w₁ /2

we look for the variation of the kinetic energy with 2

             ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)

             ΔK = ¼ I₀ w₁² -½ I₀ w₁²

             ΔK = - ¼ I₀ w₁²

the negative sign indicates that the kinetic energy decreases

We look for the change in Temperature with the expression 3

              ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce

              ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]

              ΔT =  [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]

              ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]

in this expression all the terms are contained

The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].

The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.

The given data in the problem is;

[tex]\rm \omega_1[/tex]  is the angular velocity of  disk 1

[tex]\rm \omega_2[/tex] is the  angular velocity of disk 2

r₁ is the radius of  disk 1

r₂ is the radius of  disk 2

m₁ is the mass of disk 1

m₂ is the mass of disk 2

Momentum before the collision;

[tex]\rm L_1 = I_1 \omega_1[/tex]

The moment of inertia of disc 1

[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]

The momentum gets conserved;

[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]

The change in the kinetic energy is;

[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]

The change in the energy gets converted into heat;

[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]

The change in the temperature is

[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]

The internal energy change is found by;

[tex]\rm \triangle E = mc_v dt[/tex]

[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]

Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].

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Answer:

2π/[28 x (10^-3)]

Explanation:

Angular speed : ω=2π/T

T = 28ms = 28 x (10^-3) s

Angular speed = 2π/[28 x (10^-3)]

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]

So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].