In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis

Answer:

  v = 5.88 10⁷ m / s

Explanation:

For this exercise we use the relation

          E = m c²

also indicate that all energy is converted into kinetic energy

          E = K = ½ (M-2m) v²

where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined

we substitute  

         m c² = ½ (M-2m) v²

        v² = [tex]2 \frac{m}{M+2m} \ c^2[/tex]

let's calculate

        v = [tex]\sqrt{2 \ \frac{4 \ 10^4 }{2 \ 10^6 + 2 \ 4 \ 10^4} \ (3 \ 10^8)^2}[/tex]

         v = [tex]\sqrt{ 34.615 \ 10^{14}}[/tex]

        v = 5.88 10⁷ m / s

Answer:

The solution of the given question is summarized in the below section.

Explanation:

The given values are:

Tension,

T = 100 N

mass,

m = 0.2 kg

length,

l = 0.5 m

Now,

(a)

Somewhere at bottom, string or thread breaks since string voltage seems to be the strongest around this stage.

then,

⇒  [tex]T-mg=\frac{mv^2}{l}[/tex]

or,

⇒  [tex]T=mg+\frac{mv^2}{l}[/tex]

(b)

As we know,

⇒  [tex]\frac{mv^2}{l}=T-mg[/tex]

or,

⇒  [tex]v^2=\frac{(T-mg)l}{m}[/tex]

On substituting the values, we get

⇒       [tex]=\frac{(100-0.2\times 10)0.5}{0.2}[/tex]

⇒       [tex]=\frac{49}{0.2}[/tex]

⇒    [tex]v =\sqrt{245}[/tex]

⇒       [tex]=15.65 \ m/s[/tex]

Answer:

C

Explanation:

a=f/m

10Kgm/s2/4kg

2.5m/s2

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.

Given in the question, force 10 N and mass 4.0 Kg the acceleration is,

a = 10/4 = 2.5 m/sec²

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

Answer:

C. Slippery feel

Explanation:

Answer:

Explanation:

Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.

Answer:

its in the picture hope it helps make brainlliest ty

Answer:

Resistance of the light bulb = 1.33 ohm (Approx.)

Explanation:

Given:

Current in electric bulb = 2 Amp

Voltage in battery = 1.5 Volt

Find:

Resistance of the light bulb

Computation:

Resistance = Voltage / Current

Resistance of the light bulb = Voltage in battery / Current in electric bulb

Resistance of the light bulb = 2 / 1.5

Resistance of the light bulb = 1.33 ohm (Approx.)

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

hope you can understand

Answer:

. always start on the north pole and terminate (end) on the South Pole

Explanation:

Answer:

The car will stop moving

Explanation:

If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.

Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.

Answer:

hi your question is incomplete below is the complete question

A charge of 2.15 mC is placed at each corner of a square 0.500 m on a side. Determine the direction of the force on a charge.

answer : Along the line between the charge and the center of the square outward of the center ( A )

Explanation:

The direction of the force on a charge is along the line between the charge and the center of the square outward of the center

Given that; Fnet = 3.17 * 10^-1 N ( calculated value )

The nature of the Force is repulsive in nature

attached below is a Pictorial representation of the direction of the force on a charge

Answer:

C

Explanation:

O has 8   protones,S  tiene 16, Se 34 y Te 52.

Answer:

Force = 5.22 N

Explanation:

According to Newton's Second Law of motion:

[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]

where,

m = mass of ball = 145 g = 0.145 kg

vf = final speed of ball after hit = 25 m/s

vi = initial speed of ball before hit = -  11 m/s (negative sign due to opposite direction)

t = time of contact = 1 s

Therefore,

[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]

Force = 5.22 N

Answer:98.6 degrees Fahrenheit

Explanation:

Answer:

A. 3.9 V B. 1.9 fA

Explanation:

Part A

What emf is induced in the ring as the field changes?

Express your answer to two significant figures and include the appropriate units.

The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time

ε = ΔΦ/Δt

ε = ΔABcos0°/Δt

ε = AΔB/Δt

A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T.  ΔB = B₁ - B₀  = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.

So, ε = AΔB/Δt

ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s

ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s

ε = 3.926 V

ε ≅ 3.9 V

Part B

If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.

Express your answer to two significant figures and include the appropriate units.

Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².

So, i =  ε/R

=  ε/ρl/A

= εA/ρl

= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)

= ‭15.704‬ × 10⁻⁶ V-m²/(82.9356‬ × 10⁸ Ωm²

= 0.1894 × 10⁻¹⁴ A

= 1.894 × 10⁻¹⁵ A

≅ 1.9 fA

Answer:

Work input = Work output * Work against friction is your answer so C

Explanation:

I hope this helps you :)

Answer:

A) Work input = Work output + Work against friction

Explanation:

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating  vertically too remains the same.

Therefore, T = 0.71 seconds

Answer:

a. The laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Explanation:

a. The laser tracking mechanism experiences a changing tangential velocity

This is because the tangential velocity v = rω where r = radius of disc and ω  = angular speed of discs. Since r is constant, v ∝ ω.

Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.

So, the laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.

Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.

So, The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.

Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.

So, the laser tracking mechanism experiences a non-zero tangential acceleration

Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.

Answer:

C: Ignoring

Explanation:

On edge 2021

Answer:

It's C.

Explanation:

Solution :

In the question, it is given that the collision is inelastic and the blocks stick together.

In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.

The linear momentum is given by :

[tex]$\vec p = m \vec v$[/tex] (mass x velocity)

So according to the conservation of linear momentum,

[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]

Let the velocity after the collision is [tex]$v_F$[/tex]

[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]

Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]

[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]

∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.

and  [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.

Answer:

a)     fem = - 0.0103 V,  b) the applied field is in a vertical upward direction, the induced current is clockwise.

Explanation:

a) For this exercise we use Faraday's law

         fem =  [tex]- \frac{d \phi}{dty}[/tex]

the magnetic flux is

         Ф = B. A = B A cos θ

The bold letters indicate vectors, in this case the direction of the magnetic field and the normal to the circumference is parallel therefore the angle is zero and the cos 0 = 1

         fem = - B dA / dt

the area of ​​a circle is

        A = π r²

l

et's perform the derivative

         dA / dt =π 2r [tex]\frac{dr}{dt}[/tex]

we substitute

          fem = - B 2π r [tex]\frac{dr}{dt}[/tex]

the circumference of a circle is

         L = 2π r

we substitute

          fem = - B L  ( L  )

          fem =  

Let's find the circumference for the 9 s, let's use a direct rule of proportions

If the circumference changes 13cm at t = 1. how much does it change at t=9s

         ΔL = 13cm (9s / 1s) = 117cm

the circumference that is

        L = Lo - ΔL

        L = 167 - 117

        L = 50 cm

let's reduce all magnitudes to the SI system

         L = 0.50 m

          = 0.130 m / s

calculate us

         fem = - 1.00 0.50 0.130

         fem = - 0.0103 V

b) the electromotive force induced in the opposite direction to the change of the radius and is decreasing with time, the current follows the direction of the decreased voltage therefore the current is induced in the opposite direction to the change of the magnetic flux.

If the applied field is in a vertical upward direction, the induced current is clockwise.

Answer:

The appropriate solution is "23.87 rev".

Explanation:

The given values are:

Initial angular velocity,

[tex]\omega_i=20 \ rad/s[/tex]

Final angular velocity,

[tex]\omega_f=40 \ rad/s[/tex]

Time taken,

[tex]t = 5.0 \ s[/tex]

If, α be the angular acceleration, then

⇒  [tex]\omega_f=\omega_i+\alpha t[/tex]

or,

⇒  [tex]\alpha t=\omega_f-\omega_i[/tex]

⇒  [tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

On substituting the values, we get

⇒     [tex]=\frac{40-20}{5.0}[/tex]

⇒     [tex]=\frac{20}{5.0}[/tex]

⇒     [tex]=4 \ rad/s^2[/tex]

If, ΔФ be the angular displacement, then

⇒  [tex]\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2[/tex]

On substituting the values, we get

⇒        [tex]=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2][/tex]

⇒        [tex]=100+50[/tex]

⇒        [tex]=150[/tex]

On converting it into "rev", we get

⇒  [tex]\Delta \theta=(\frac{150}{2 \pi} )[/tex]

⇒        [tex]=23.87 \ rev[/tex]

Answer:

Soil fertility can be further improved by incorporating cover crops that add organic matter to the soil, which leads to improved soil structure and promotes a healthy, fertile soil; by using green manure or growing legumes to fix nitrogen from the air through the process of biological nitrogen fixation; by micro-dose ........

Answer:

Continuous cultivation of water intensive crops like sugarcane and rice has resulted in their water tables running low at 149 percent. Alternating crops with millet, which consumes less water and has a high nutritional value, can increase the soil fertility and balance nutrients present in it.Apr 19, 2019

Explanation:

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hope my answer is helpful

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + [tex]\frac{1}{2}[/tex] ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

Δr = 0.42368 m

v = f × λ

f = ( 2m + [tex]\frac{1}{2}[/tex])v / Δr

m = 0, 1, 2, 3, ....

Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + [tex]\frac{1}{2}[/tex])v / Δr

f = [tex]\frac{1}{2}[/tex](v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = [tex]\frac{1}{2}[/tex](343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object.

Momentum is force or speed of movement.

Velocity defines the path of the motion of the frame or the object

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

This question is incomplete, the complete question is;

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]

hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1

we solve for W

W = hc/λ - [tex]K_{max[/tex]

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] -  3.28 × 10⁻¹⁹

W =  8.99457 × 10⁻¹⁹ -   3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + [tex]K'_{max[/tex]

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ /  1.227457 × 10⁻¹⁸

λ' =  1.619 × 10⁻⁷ m

λ' =  161.9 nm

Therefore, the required wavelength of light is 161.9 nm

Answer:

b

Explanation:

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

[tex]F = \frac{Kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.