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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:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2

Answer:

b

Explanation:

f1=-f2 that could be thank u

Answer:

4.8 m/s

Explanation:

Given: angular velocity of wheel ω = 2.5 rev/sec

radius r = 30 cm

length of arrow = 24 cm

For arrow to pass through spinning ring it has to pass between any two spokes of the wheel.

angle between two spokes = π/4

time taken by a spook to reach the position of adjacent spoke t =θ/ω

=  π/4/(2.5×2π) = 1/20 sec

for the arrow to pass through the spokes of the wheel it should take time t <1/20 sec to pass through the wheel

a) therefore, minimum speed = (24/100)/(1/20) = 4.8 m/s

The average temperature of the air in the room at 20 min is 23°C.

Temperature is the degree of hotness or coldness of the object.

A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.

The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.

Thus, the average temperature of the air in the room at 20 min is 23°C.

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Answer:

5.45 moles

Explanation:

The chemical balanced equation of this reaction is;

3Si + 2N2 → Si3N4

From the balanced equation, we can see that that 3 moles of Si reacts with 2 moles of N2 to produce 1 mole of Si3N4.

Thus imies that the molar ration of Si to N2 is 3:2.

Now, we are told that 16.35 moles of Si reacts with 11.26 moles of N2.

16.35

Thus, using the ratio 3:2, we can say that moles of 16.35 miles of Si will react completely with (16.35 × ⅔) = 10.9 moles of N2.

Remaining N2 = 11.26 - 10.9 = 0.36 will be the excess.

From our balanced chemical equation, we saw that;

3 moles of Si produced 1 mole of Si3N4.

Thus; 16.35 moles of Si will produce;

no. of moles of Si3N4 produced = (1 × 16.35)/(3.0) = 5.45 moles

Answer: 84 %

Explanation:

Answer:

huh

Explanation:

Answer:

Amplitude

Explanation:

The loudness of a sound depends on the amplitude of vibration producing the sound

Answer:

0.706

Explanation:

Since the other astronaut measures a longer time, this is a time dilation problem. So, our equation for time dilation is given by

T = T₀/√(1 - β²) where T = period on passing space ship = 31.87 s, T₀ = period on other space vehicle = proper time = 22.58 s and β = relative velocity of between the two observers.

T = T₀/√(1 - β²)

√(1 - β²) = T₀/T

squaring both sides, we have

[√(1 - β²)]² = (T₀/T)²

1 - β² = (T₀/T)²

β² = 1 - (T₀/T)²

taking square root of both sides, we have

√β² = √[1 - (T₀/T)²]

β = √[1 - (T₀/T)²]

substituting the values of the variables into the equation, we have

β = √[1 - (22.58 s/31.87 s)²]

β = √[1 - (0.7085)²]

β = √[1 - 0.502]

β = √0.498

β = 0.706

Answer:

83.3 m

Explanation:

From the question,

Applying on of the formula of velocity

V = d/t.................. Equation 1

Where V = average velocity, d = distance, t = time.

make d the subject of the equation

d = V×t................. Equation 2

Given: V = 9.8 m/s, t = 8.5 s

Substitute these values into equation 2

d = 9.8×8.5

d = 83.3 m

Hence the total distance covered by the rabbit is 83.3 m

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex]     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

Answer:

[tex]p =-1.03[/tex]

Explanation:

From the question we are told that:

Focal length of lens [tex]f=-0.304 m[/tex]

Image distance [tex]q=0.237 m[/tex]

Height of image [tex]H_i=0.343[/tex]  

Generally the lens equation is mathematically given by

[tex]\frac{1}{f} = \frac{1}{q} - \frac{1}{p}[/tex]

Where [tex]p[/tex] is Subject

[tex]p = \frac{(qf) }{(f - q)}[/tex]

[tex]p = \frac{(-0.237)(-0.304)) }{((-0.304) - (0.237))}[/tex]

[tex]p =-1.03[/tex]

Therefore the distance between  the person and the car is

[tex]p =-1.03[/tex]

Answer:

Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Explanation:

If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.

Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Answer:

I = 0.65 kgm²

Explanation:

Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.

T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.

Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation

So, T = 2.186 s

We now find I by making it subject of the formula in the equation for T.

So,

T = 2π√(I/mgh)

dividing both sides by 2π, we have

T/2π = √(I/mgh)

squaring both sides, we have

(T/2π)² = [√(I/mgh)]²

T²/4π² = I/mgh

multiplying both sides by mgh, we have

T²mgh/4π² = I

I = T²mgh/4π²

substituting the values of the variables into the equation, we have

I = T²mgh/4π²

I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I =  29.539 kgm²/4π²

I = 0.748 kgm²

Now I = I' + mh²  (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.

So, I' = I - mh²

Substituting the values of the variables into the equation, we have

I' = I - mh²

I' = 0.748 kgm² - 4.07 kg × (0.155 m)²

I' = 0.748 kgm² - 4.07 kg × 0.02403 m²

I' = 0.748 kgm² - 0.098 kgm²

I = 0.65 kgm²

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:

i dont know but i should know try g o o g l e

Explanation:

Answer:

The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].

Explanation:

The work done is given as in terms of

[tex]W=\Delta TE[/tex]

Where ΔTE is the change in total energy.

This is given as

[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]

Rearranging it in  terms of K_x gives

[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]

Answer:

the decay of uranium ending in lead,  of potassium (40K) that becomes argon,  the decay of rubidium

Explanation:

For the radioactive dating process, a material is needed that has a known average life time and that we can know the amount of material at a given moment,

In the case of carbon 14 (14C), living beings stop capturing it from the air and plants when they die, so knowing the amount they currently have, it is possible to calculate the time in which they stopped absorbing, but the life time average is 5730 years, the maximum time that can be used is up to about 10 average visa times

To analyze extra samples have high half-life times

* the decay of uranium ending in lead

* the decay of potassium (40K) that becomes argon T1 / 2 = 1,251 10⁹ years

* the decay of rubidium (87Ru) which becomes strontium T1 / 2 = 4.92 10¹⁰ years

Answer:

The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.

Explanation:

I hope this is what you were looking for?!

Hope this helps!

Have a great day!

-Hailey!

The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.

A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.

Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

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Answer:

   f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

        ∑ F = ma

where the acceleration is

         a = [tex]\frac{d^2 y}{dt^2 }[/tex]

        B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

           B = W

In this frame of reference, the variable y'  when it is oscillating is positive and negative, therefore Newton's equation remains

         B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]

the thrust is given by the Archimedes relation

         B = ρ_liquid g V_liquid

the volume is

        V = π r² y'

we substitute

          - ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

          [tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]

this differential equation has a solution of type

         y = A cos (wt + Ф)

where

         w² = ρ_liquid g π r² /m

angular velocity and frequency are related

         w = 2π f

we substitute

          4π² f² = ρ_liquid g π r² / m

          f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]

calculate

         f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]

         f = 5.3 Hz

Answer:

  K_{total} = 8.91 J

Explanation:

In this exercise you are asked to find the kinetic energy of the can of coca-cola

         K_total = K_ {Translation} + K_ {rotation}

the translational kinetic energy is

         K_ {translation} = ½ m v²

the kinetic energy of rotation is

         K_ {rotation} = ½ I w²

The moment of inertia of a cylinder is

           I = ½ m r²

we substitute

          K_ {total} = ½ m v² + ½ (½ m r²) w²

angular and linear velocity are related

          v = w r

we substitute

          K_ {total} = ½ m v² + ¼ m r² v² / r²

          K_ {total} = m v² (½ + ¼)

          K_ {total} = ¾ m v²

let's calculate

          K_ {total} = ¾ 0.33 6.00²

           K_{total} = 8.91 J

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

C

Explanation:

Primacy means being first or important so thats not an important facial display as the others.

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

Answer:

a. Speed = 1.6 m/s

b. Amplitude = 0.3 m

c. Speed = 1.6 m/s

Amplitude = 0.15 m

Explanation:

a.

The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:

frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz

The wavelength of the wave is the distance between consecutive crests of wave. Therefore,

Wavelength = 6.4 m

Now, the speed of the wave is given as:

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

b.

Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:

Amplitude = (0.5)(0.6 m)

Amplitude = 0.3 m

c.

frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

Amplitude = (0.5)(0.3 m)

Amplitude = 0.15 m

Answer:

false

Explanation:

Answer:

Truck [tex]\dfrac{g}{10}[/tex]

Road [tex]-\dfrac{g}{10}[/tex]

Explanation:

[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]

[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]

Frictional force is given by

[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]

Net acceleration is given by

[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]

The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.

Answer: I think B.)

Explanation: