You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you slowly pull down with a spring scale so the block is now at position y bottom, below the equilibrium position y0 where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
a. Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is a positive direction.
b. Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
c. What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
d. How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.

Answer:

4.57x10^-6

Explanation:

So first you'll need to find the current, which is

I = voltage / resistance

which equals 3.43

Then just plug your numbers into the magnetic field equation to get your final answer

4.57x10^-6

Also it was right on Acellus lol

Hope this helps :)

The magnetic field 0.150 m from the wire is 2.86 x 10⁻⁵ T.

The magnetic field created by a current-carrying wire is inversely proportional to the distance from the wire. The closer you get to the wire, the stronger the magnetic field. The distance from the wire is 0.150 m. The voltage applied to the wire is 12.0 V and the resistance of the wire is 3.50 ohms.

We can use the following equation to calculate the magnetic field:

B = μ0 * I / 2πr

B = magnetic field (T)

μ0 = permeability of free space (4π x 10^-7 T m/A)

I = current (A)

r = distance from wire (m)

Plugging in the values, we get:

B = (4π x 10⁻⁷ T m/A) * (12.0 V / 3.50 ohms) / 2π * 0.150 m

B = 2.86 x 10⁻⁵ T.

As a result, the magnetic field 0.150 m away from the wire is 2.86 x 10⁻⁵ T.

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Answer:

2.3 m/s

Explanation:

A fisherman notices that one wave passes the bow of his anchored boat every 3 seconds. The speed of the wave traveling is 2.83 m/s. The correct option is 1.

Velocity is a vector expression of an object's or particle's displacement with respect to time. The meter per second (m/s) is the standard unit of velocity magnitude (also known as speed). The meter per second (m/s) is a unit of velocity magnitude (also known as speed).

Speed is the rate at which an object moves along a path in time, whereas velocity is the rate and direction of movement. In other words, velocity is a vector, whereas speed is a scalar value.

Frequency = 1/time

Time = 3 seconds

λ = 8.5 meters.

Putting the value in the formula:

v = f x λ

v = 1/3 x 8.5 = 2.83 m/s

Therefore, the speed of the wave traveling is 2.83 m/s. The correct option is 1.

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The question is incomplete. Your most probably complete question is given below:

1.)2.83m/s

2.)0.283m/s

3.)28.3m/s

Answer:

kg m2/s

Explanation:

I think it is :)

The SI unit of angular momentum is kg·m²/s, reflecting an object's rotational motion dynamics. Fundamental to rotational dynamics, angular momentum is calculated based on an object's moment of inertia and angular velocity, adhering to the dimensions ML²T-¹.

The SI unit of angular momentum is the kilogram meter squared per second (kg·m²/s). Angular momentum, denoted by the symbols “l” for an individual particle and “L” for a system of particles or a rigid body, represents the rotational equivalent of linear momentum. It is fundamentally connected to the concepts of moment of inertia and angular velocity, where the angular momentum (“l” or “L”) can be calculated by the product of these two quantities. This unit emphasizes the momentum of an object in rotational motion about an axis and is derived from taking the product of the object's moment of inertia and its angular velocity.

The calculation of angular momentum involves factors such as the mass (m) of the particle, its velocity (v) perpendicular to the line joining it to the axis of rotation, and its distance (r) from the axis. This relationship is encapsulated in the formula L=mvr, highlighting the linear momentum's component (mv) and its lever arm distance (r) from the rotation axis. The SI units and dimensions (ML²T-¹) of angular momentum validate its role in describing the rotational dynamics of objects.

Answer:

a) [tex]f=0.64 Hz[/tex]

b) [tex]\lambda=62.5 cm[/tex]

c) [tex]y_{m}=5 cm[/tex]

d) [tex]k=0.1 cm^{-1}[/tex]

e) [tex]\omega=4 s^{-1}[/tex]

g) [tex]T=0.064 N[/tex]    

Explanation:

We know that the wave equation is of the form:

[tex]y(x,t) = y_{m}sin(kx \pm \omega t)[/tex]

Comparing with the equation of the sinusoidal wave [tex](y=(5.0cm)sin[1.0−(4.0s^{-1})t])[/tex] we will have:

[tex]\omega = 4 s^{-1}[/tex]

a) ω is the angular frequency and it can writes in terms of frequency as:

[tex]\omega = 2\pi f[/tex] , where f is the frequency.

[tex]f=\frac{\omega}{2\pi}[/tex]

[tex]f=0.64 Hz[/tex]

b) Let's recall that the speed of the wave is the product between the wave length and the frequency, so we have:

[tex]v=\lambda f[/tex]

[tex]\lambda=\frac{v}{f}[/tex]

v is 40 cm/s

[tex]\lambda=\frac{40}{0.64}[/tex]

[tex]\lambda=62.5 cm[/tex]

If we compare each equation we can find y(m), k and ω:

c) [tex]y_{m}=5 cm[/tex]

d) [tex]kx=1[/tex] and we know that x = 10 cm, so:

[tex]k=\frac{1}{x}=\frac{1}{10}[/tex]

[tex]k=0.1 cm^{-1}[/tex]

e) [tex]\omega=4 s^{-1}[/tex]

f) The minus sign in front of the angular frequency in the equation is the correct choice, just by comparing.

g) We have to use the equation of the speed in terms of tension.

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

[tex]T=\mu*v^{2}[/tex]

[tex]T=4*40^{2}=6400 g*cm*s^{-2}[/tex]    

[tex]T=0.064 N[/tex]    

I hope it helps you!

(a) The frequency of the wave is 6.37 Hz and (b) the wavelength is 2π cm. (c) The values for y_m, k, and ω are 5.0 cm, (d) 1 cm^-1, and (e) 4.0 s^-1 respectively. (f) The correct choice of sign in front of ω is positive. (g) The tension in the string is 12533.5 g*cm²/s².

(a) To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the velocity, and λ is the wavelength. In this case, v = 40 cm/s and we need to find λ. Since y(x,t) = y_m sin(kx ± ωt), we can compare it to the given equation y=(5.0cm)sin[1.0−(4.0s−1)t] to find k and ω. From the comparison, we know k = 1 cm^-1 and ω = 4.0 s^-1. Therefore, the wavelength is given by λ = 2π/k, and plugging in the values, we get λ = 2π cm. The frequency can then be calculated as f = v/λ = 40 cm/s / 2π cm = 6.37 Hz.

(b) the wavelength is 2π cm.

(c) y_m is the amplitude of the wave, which is 5.0 cm.

(d) k is the wave number, which is 1 cm^-1.

(e) ω is the angular frequency, which is 4.0 s^-1.

(f) The correct choice of sign in front of ω depends on the direction of wave propagation. If the wave is traveling in the positive x-direction, the sign should be positive, so ωt is correct.

(g) To find the tension in the string, we can use the formula T = λ²μf², where T is the tension, λ is the wavelength, μ is the linear density, and f is the frequency. Plugging in the values, we get T = (2π cm)² * (4.0 g/cm) * (6.37 Hz)² = 12533.5 g*cm²/s².

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Final answer:

The shortest time it takes for the child to catch up with the horse can be found by analyzing the angular motion of the merry-go-round and the child. Using the given values of angular speed and angular acceleration, we can set up an equation and solve for time using the quadratic formula.

Explanation:

To find the shortest time it takes for the child to catch up with the horse, we need to analyze the angular motion of the merry-go-round and the child.

Given that the child's angular speed is 0.233 rad/s and the merry-go-round starts rotating with an angular acceleration of 0.0136 rad/s^2, we can use the equation:
Δθ = ω_0 * t + (1/2) * α * t^2
where Δθ is the angle the child needs to catch up with the horse (π/2 radians in this case), ω_0 is the initial angular speed of the merry-go-round, α is the angular acceleration, and t is the time.

Plugging in the values, we have:
π/2 = 0.233 * t + (1/2) * 0.0136 * t^2

This equation is quadratic in form, so we can solve it using the quadratic formula. The positive root will give us the time it takes for the child to catch up with the horse.

Answer:

Gas exchange in fish is by counter current exchange

Explanation:

The gills located at the pharynx of a fish is a very important respiratory organ.Oxygen and carbon dioxide are the substance of exchange in fishes during respiration.During exchange a fish takes in a needed volume of water through the mouth,then moves it through the gills which aids in repleting oxgen poor water out through various opening and also helps in replenishing the blood capillaries flowing in the opposite direction with oxygen

Answer:

By using countercurrent flow principle of water and blood to exchange oxygen.

Explanation:

Fish use a specialized organ called gills to carry out gas exchange.

Gills have a lot of folds, maximizing their surface area and maximising the efficiency of gas exchange. The gill filaments have protrusions called gill lamellae.

One of the ways in which gas exchange is carried out efficiently is by the countercurrent flow principle, which simply means that water and blood are flowing in different directions. The water that passes over the gill lamellae flows in the opposite direction to the blood within the gill lamellae.

Answer:

Why don't you search for the formula, this one is really basic. The answer is below

Explanation: The formula t= d/s or d= st or s= d/t in this problem we use:

t = d/s

t is time

d is distance and s is speed

so just plug in the data which are given in the question,

185 km is the distance,d

95 km/h is the speed, s

It tell you to calculate how long means t= 185/95 ≈ 1.947 or 2 hours

Final answer:

To calculate the time it will take to drive 185 km at an average speed of 95 km/h, you divide the distance by the speed to get approximately 1 hour and 57 minutes.

Explanation:

The question asks us to calculate the time it will take to travel to your grandparents' house at a constant speed. This is a classic speed, distance, and time problem that can be solved using the formula: time = distance \/ average speed. To find the time it will take to reach the grandparents' house, we divide the distance by the average speed.

Step-by-Step Calculation:

Distance to grandparents' house = 185 km.

Average speed = 95 km/h.

Time = Distance \/ Average Speed = 185 km \/ 95 km/h.

Time = 1.9474 hours.

To convert hours to minutes, we multiply by 60 (since each hour has 60 minutes).

Time = 1.9474 hours × 60 minutes/hour = 116.842 minutes, which is approximately 1 hour and 57 minutes.

Therefore, it will take nearly 1 hour and 57 minutes to get to your grandparents' house if you drive at an average speed of 95 km/h.

Final answer:

A car with tires of radius 0.280 m decelerating at 7.00 m/s² has an angular acceleration of 25 rad/s². To find the number of revolutions before stopping, kinematic equations for rotational motion are used with an initial angular velocity of 95.0 rad/s.

Explanation:

A car decelerating at 7.00 m/s² with tires of radius 0.280 m requires an understanding of the relationship between linear acceleration and angular acceleration. The formula linking these two is α = a/r, where α is the angular acceleration, a is the linear acceleration, and r is the radius of the tire. Using these values, the angular acceleration is calculated to be 25 rad/s².

For part (b), to find out how many revolutions the tires make before coming to rest, we need to use kinematic equations for rotational motion. Initially, we have an angular velocity (ω) of 95.0 rad/s and we want to find the total angle (θ) covered by the tires. Using the equation θ = (ω0²)/(2*α), we calculate the angle in radians and then convert this to revolutions by dividing by 2π. After finding the revolutions, the car's tires will have stopped rotating completely.

Answer:

Yes it is possible

Explanation:

These are the workhorses of the living world, speeding up chemical reactions so that processes that would otherwise take thousands of years happen inside living cells in seconds. How they achieve this speed-up – often more than a trillion-fold – has long been an enigma. But now, research by Judith Klinman at the University of California, Berkeley and Nigel Scrutton at the University of Manchester (among others) has shown that enzymes can employ a weird quantum trick called tunnelling. Simply put, the enzyme encourages a process whereby electrons and protons vanish from one position in a biochemical and instantly rematerialise in another, without visiting any of the in-between places – a kind of teleportation.

Final answer:

Quantum tunneling is indeed a process that can be used by enzymes in our bodies to facilitate biochemical reactions by allowing particles to penetrate potential energy barriers, significantly affecting reaction rates.

Explanation:

Yes, it is possible that the enzymes in our bodies use quantum tunneling. Quantum tunneling is a phenomenon whereby particles penetrate a potential energy barrier despite having total energy less than the height of the barrier, which defies the principles of classical mechanics. This process was first analyzed by Friedrich Hund in 1927 and later used by George Gamow to explain alpha decay of atomic nuclei as a quantum-tunneling phenomenon.

In biological systems, quantum tunneling allows for the transfer of protons or electrons during enzyme-catalyzed reactions, significantly affecting the reaction rate by providing an alternate, low-energy pathway for the reaction to occur. Thus, enzymes can use quantum tunneling to speed up biochemical reactions that would otherwise occur much more slowly. This principle is vital in fields like biochemistry and molecular biology, providing a deeper understanding of enzyme kinetics and mechanism.

Answer:

[tex]v_0=17.3m/s[/tex]

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

[tex]E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2[/tex]

Since we only care about the velocity [tex]v_0[/tex], we can keep only the second and third parts of the equation and solve:

[tex]mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s[/tex]

So, the speed of the ball just after the launch is 17.3m/s.

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

Answer:

40 m

Explanation:

English Translation for the question

Suppose Lisa needs to find a nearby bakery because she needs to improve her time to 80s. How far should the bakery be if she moves at a speed of 0.5m/s?

Speed = (Distance/time)

Speed = 0.5 m/s

Distance = how far away the bakery should be = d = ?

Time = 80 s

0.5 = (d/80)

d = 0.5×80 = 40 m

Hope this Helps!!!

In this Calculating Distance question, Para determinar la distancia a la que debe estar la panadería, podemos utilizar la fórmula de velocidad promedio. Reemplazando los valores conocidos, la panadería debe estar a una distancia de 40 metros.

Para calcular la distancia a la que debería estar la panadería, podemos utilizar la fórmula de velocidad promedio:

Velocidad promedio = Distancia / Tiempo

En este caso, la velocidad es de 0.5 m/s y queremos calcular la distancia. Dado que el tiempo es de 80 segundos (el cual debe convertirse a minutos), podemos reorganizar la fórmula para despejar la distancia:

Distancia = Velocidad promedio x Tiempo

Reemplazando los valores conocidos, tenemos:

Distancia = 0.5 m/s x 80 s = 40 metros

Por lo tanto, la panadería debe estar a una distancia de 40 metros para que Lisa mejore su tiempo a 80s.

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Answer:

Part(a): The expression for the velocity of the billiard ball is [tex]\bf{v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}}[/tex]

Part(b): The value of the velocity of the billiard ball is [tex]\bf{5.57~m/s}[/tex].

Part(c): The expression for the velocity of the cue ball is [tex]\bf{v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}}[/tex]

Part(d): The value of the velocity of the cue ball is [tex]\bf{1.93~m/s}[/tex].

Explanation:

Given:

The mass of the cue ball, [tex]m_{1} = 0.34~kg[/tex].

The mass of the billiard ball, [tex]m_{2} = 0.575~kg[/tex].

The initial velocity of the cue ball, [tex]u_{1} = 7.5~m/s[/tex]

The initial velocity of the billiard ball, [tex]u_{2} = 0[/tex]

(a)

Consider the final velocity of the cue ball be [tex]v_{1}[/tex] and the final velocity of the billiard ball be [tex]v_{2}[/tex].

From the conservation of linear momentum , we can write

[tex]~~~~&& m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\&or,& m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

From the conservation of energy, we can write

[tex]~~~&& \dfrac{1}{2}m_{1}u_{1}^{2} + \dfrac{1}{2}m_{2}u_{2}^{2} = \dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2}\\&or,& \dfrac{1}{2}m_{1}(u_{1}^{2} - v_{1}^{2}) = \dfrac{1}{2}m_{2}(v_{2}^{2} - u_{2}^{2})~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Dividing equation (1) by equation (2), we have

[tex]~~~&& \dfrac{m_{1}(u_{1} - v_{1}) }{m_{1}(u_{1}^{2} - v_{1}^{2})} = \dfrac{m_{2}(v_{2} - u_{2})}{m_{2}(v_{2}^{2} - u_{2}^{2})}\\&or,& u_{1} + v_{1} = u_{2} + v_{2}~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]

Rearranging equation (3) for [tex]v_{1}[/tex], we have

[tex]v_{1} = u_{2} + v_{2} - u_{1}~~~~~~~~~~~~~~~~~~~~(4)[/tex]

Substitute equation (4) in equation (1), we can write

[tex]v_{2} = \dfrac{2m_{1}u_{1}}{m_{1}+m_{2}} + \dfrac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(5)[/tex]

Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{2f}[/tex] for [tex]v_{2}[/tex] in equation (5), we have

[tex]v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(6)[/tex]

(b)

Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (6), we have

[tex]v_{2f} &=& \dfrac{2(0.34~kg)(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& 5.57~m/s[/tex]

(c)

Rearranging equation(3) for [tex]v_{2}[/tex], we have

[tex]v_{2} = u_{1} + v_{1} – u_{2}~~~~~~~~~~~~~~~~~~~~(7)[/tex]

Substitute equation (7) in equation (1), we can write

[tex]v_{1} = \dfrac{2m_{2}u_{2}}{m_{1}+m_{2}} + \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(8)[/tex]

Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{1f}[/tex] for [tex]v_{1}[/tex] in equation (8), we have

[tex]v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(9)[/tex]

(d)

Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (9), we have

[tex]v_{1f} &=& \dfrac{(0.34 - 0.575)~kg(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& -1.93~m/s[/tex]

Negative sign indicates that the cue ball will bounce back.

The expressions for the horizontal component of the billiard ball's velocity after the collision and the cue ball's velocity after the collision are given. The velocities can be calculated using the provided equations and given values.

a) The expression for the horizontal component of the billiard ball's velocity, vr after the collision is given by:

vr = (m1 - m2) * (v1 / (m1 + m2))

b) Substituting the given values into the equation, we find that the horizontal component of the billiard ball's velocity after the collision is approximately 4.02 m/s.

c) The expression for the horizontal component of the cue ball's velocity, vr after the collision is given by:

vr = (2 * m2 * v1) / (m1 + m2)

d) Substituting the given values into the equation, we find that the horizontal component of the cue ball's final velocity is approximately 2.48 m/s.

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Answer:

[tex]2\Omega[/tex]

Explanation:

(Assuming the cell in the circuit has 0 internal resistance)

Ohm's Law is given as:

[tex]V=IR[/tex]

Voltage is Current multiplied by Resistance.

We can rearrange this formula to give us:

[tex]R=\frac{V}{I}[/tex]

Now we can plug in our values

[tex]R=\frac{6}{3}=2\Omega[/tex]

I= 3A

V= 6V

R=?

V=IR

R=V/I

R=6/3

Therefore R = 2ohms

Hope this will help u mate :)

Answer:

The added mass m= 0.7kg

Explanation:

This problem bothers on the simple harmonic motion of a spiral spring

We know that the period of a simple harmonic motion of a spring is given as

T=2π√m/k

We need to solve first for the spring constant k

Given data

Mass m =230g - - - - - kg

=230/1000= 0.230kg

Period T = 1sec

Substituting we have

1= 2*3.142√0.230/k

1=6.284√0.230/k

1/6.284=√0.230/k

Square both sides

(0.159)²=0.230/k

0.025=0.230/k

k=0.230/0.025

k= 9.2N/m

Now we find that the period of oscillation is 2 after adding mass m to 230g.. Let's solve for the new mass

Using the formula for the period T=2π√m/k

2=2*3.142√m/9.2

2=6.284√m/9.2

2/6.284=√m/9.2

Square both sides

(0.318)²=m/9.2

0.10=m/9.2

m= 0.930kg

Therefore the added mass is

0.930kg-0.230kg

The added mass m= 0.7kg

Answer: Technical B is right.

Explanation:

The first systems injected air very close to the engine, either in the cylinder head's exhaust ports or in the exhaust manifold

However, the extra heat of recombustion, particularly with an excessively rich exhaust caused by misfiring or a maladjusted carburetor, tended to damage exhaust valves and could even be seen to cause the exhaust manifold to incandesce.

Answer: the nasal conchae also known as the turbinates

Explanation:

Answer:

the corect answer it b and c

Explanation:

b. the density of lines shows the strength of the force.

c. the arrows on the lines of force show which way a posative object will move.

youre welcome

Answer:

they are both equal

Explanation:

check the laws of reflection

Answer:

a) archimedes principle

The operation of a hydraulic jack is an application of  Pascal's principle. Hence, option (c) is correct.

Pascal's principle in fluid mechanics asserts that a pressure change in one component of a fluid at rest in a closed container is transferred without loss to every portion of the fluid and to the container walls.

The force multiplied by the surface area on which it acts produces pressure. Pascal's principle states that a pressure rise on one piston in a hydraulic system causes an equivalent increase in pressure on another piston in the system.

Even though the pressure on the second piston is the same as that on the first piston, the force acting on it is 10 times more if its area is 10 times greater than the first piston's.

The hydraulic press, which is based on Pascal's concept and is employed in systems like hydraulic jacks , is a good example of this effect.

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Answer:

To drive at high speed around an unbanked horizontal curve on the earth.

Explanation:

Garvitational pull on on eart is 6 times more than on the moon. On. Earth, the vehicle will have more grip on the surface due to its weight. This grip will reduce its tendency to skid off the horizontal curve when compared to driving in the same unbanked horizontal curve on the moon.

Final answer:

Driving at high speed around an unbanked horizontal curve would be easier on Earth compared to the moon, because Earth's greater gravity provides a higher force of static friction, reducing the risk of slipping at high speeds.

Explanation:

When deciding whether it would be easier to drive at high speed around an unbanked horizontal curve on the moon or the earth, with other things being equal, we need to consider the force of static friction and the acceleration required to maintain uniform circular motion. The acceleration in question is given by the equation |a| = |v|²/r, where |a| is the magnitude of acceleration, |v| is the speed of the car, and r is the radius of the circular path.

On the moon, due to its lower gravity, the force of static friction is less than it is on earth. Despite that, it would be more challenging to drive at high speed on the moon because a lower force of static friction means a lower threshold for slipping. On earth, the higher gravity increases the maximum force of static friction, allowing for a presumably safer high-speed turn, assuming no additional adverse conditions like rain or mud. Therefore, b. To drive at high speed around an unbanked horizontal curve on the earth would be easier for the given scenario.

Answer:

The magnetic field is [tex]2 \times 10^{-4}[/tex] T

Explanation:

Given:

Velocity of electron [tex]v = 5 \times 10^{7} \frac{m}{s}[/tex]

Electric field [tex]E = 10^{4} \frac{V}{m}[/tex]

The force on electron in magnetic field is given by,

 [tex]F = qvB \sin \theta[/tex]                      ......(1)

The force on electron in electric field is given by,

 [tex]F = qE[/tex]                               ......(2)

Compare both equation,

   [tex]qE = qvB \sin \theta[/tex]

Here [tex]\sin \theta = 1[/tex]

  [tex]E = vB[/tex]

  [tex]B= \frac{E}{v}[/tex]

  [tex]B = \frac{10^{4} }{5 \times 10^{7} }[/tex]

  [tex]B = 2 \times 10^{-4}[/tex] T

Therefore, the magnetic field is [tex]2 \times 10^{-4}[/tex] T

Answer:

1.contact force

2. Non contact

Explanation:

1. Because bodies are in contact

2.Bodies are not in contact.

Answer:

1. contact

2. non contact

Explanation:

just did it

Answer:

No, since they don't have the necessary mass and a high temperature in their core.

Explanation:

There are two forces that play an important role in the stars: the force of gravity in the inward direction due to stars' own mass and the radiation pressure in the upward direction as a consequence of the nuclear reaction in their core.

The superficial layers of the stars compress the core as an effect of their own gravity. Therefore, atoms will be closer to each other in the core, allowing them to combine, increasing the density and temperature.  

A nuclear reaction occurs when light elements combine into heavier elements (that is known as nucleosynthesis). To get fusion reactions that generate heavier elements than carbon high temperature is necessary, which can be gotten by a more massive star for what was already explained in the first paragraph.

Answer:

Technician B is correct

Explanation:

Electrical Switches makes or brakes the flow of signal (current or voltage) in a circuit.

The Technician B that says the switch contacts are closed is correct because it is until when the circuit is continues that voltage flows across the switch and voltmeter will read the amount of voltage present in the circuit. This simply means closed circuit.

Technician A is not correct because when the switch contacts are open, voltage does not flow and no reading will be measured.

Answer:

Technician B is correct

Explanation:

When a circuit is open there is no energy that allows the flux of charge in the cable. Voltage determines the capability of doing a work over charges to generate a charge flux. An open circuit is not able to generate a current in the circuit and concequently, ther is no voltage in the circuit.

Hence, a measurement of 12V in a circuit means that the circuit is closed, store energy and can generate a flux of charge.

Technician B is correct.

Answer:

1 -f,2 -b,3-e,4 -g,5-a,6 - d, 7 - h, 8 - c

Explanation:

1.  developed geocentric theory  - Aristotle

2. developed heliocentric theory - Copernicus in 1543

3. founded nursing profession   -  Florence Nightingale

4. invented barometer  -   Evangelista Torricelli in 1643

5. "Father of Modern Science" -  Galileo Galilei

6.law of universal gravitation -Sir Isaac Newton

7. examined the inner workings of the human body - Vesalius

8. developed metric temperature scale  -  Andres Celsius in 1742

Explanation:

Given

Force (F) = 500 N

Distance (d) = 2m

Now

Work done = F * d

= 500 * 2

= 1000 joule

Therefore 1000 joule of work is done.

Hope this helps.

The work done by a football player leaping 2 m into the air with a force of 500 N is calculated using the equation W = F * d, resulting in 1000 joules of work.

When calculating work done on an object, it's essential to understand that work (W) is defined as the product of the force (F) applied to an object and the distance (d) that the object moves in the direction of the force. The equation for work is:

W = F * d,

where work (W) is measured in joules (J), force (F) in newtons (N), and distance (d) in meters (m).

For a football player leaping vertically into the air with a force of 500 N over a distance of 2 m:

W = (500 N) * (2 m),

W = 1000 J.

The work done by the football player in leaping 2 meters into the air is therefore 1000 joules.

Answer:the resistance decrease

Explanation:

Answer:

0.635 Ω

Explanation:

Using

E-E' = IR.................... Equation 1

Where E = total Emf of the of the batteries, E' = back emf of the motor, I = current of the motor, R = combined resistance of the battery and the motor.

Make R the subject of the equation

R = E-E'/I.............. Equation 2

Given: E = 6.3 V, E' = 2.1 V, I = 3.1 A.

Substitute into equation 2

R = (6.3-2.1)/3.1

R = 4.2/3.1

R = 1.355 Ω

Since the motor and the batteries are connected in series,

R = R'+r'....................... Equation 3

Where R' = Resistance of the motor, r' = resistance of the batteries.

make R' the subject of the equation

R' = R-r'...................... Equation 4

Given: R = 1.355 Ω, r' = 0.18×4 (four batteries) = 0.72 Ω

Substitute into equation 4

R' = 1.355-0.72

R' = 0.635 Ω

Hence the resistance of the motor = 0.635 Ω

To find the motor's resistance, subtract the back emf from the total emf to get the effective voltage, then divide by the current and subtract the total internal resistance of the batteries. The resistance of the motor is found to be 1.35 Ω.

The question involves a toy car motor that is powered by four batteries producing a total electromotive force (emf) of 6.3 V, with each battery having an internal resistance of 0.18 Ω. Upon running, the motor develops a 2.1 V back emf and draws a current of 3.1 A. To find the resistance of the motor, one can use the net emf equation and Ohm's law.

The total emf minus the back emf is the effective voltage across the motor. Thus, the effective voltage (Veff) is 6.3 V - 2.1 V = 4.2 V. The total internal resistance (Rint) in series is 4 batteries × 0.18 Ω = 0.72 Ω. Applying Ohm's law, V = IR where I is the current and R is the resistance, we solve for the resistance (Rmotor) of the motor:

Veff = I(Rmotor + Rint)
Rmotor = Veff/I - Rint
Rmotor = 4.2 V / 3.1 A - 0.72 Ω
Rmotor = 1.35 Ω

Therefore, the resistance of the motor is 1.35 Ω.

Answer:

The radio booth is 0.993 meters above the field.

Explanation:

The pendulum covers 10 complete oscillations to take 20 s. We need to find the height above the radio booth. The time period of the pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\L=(\dfrac{T}{2\pi })^2g\\\\L=(\dfrac{(20/10)}{2\pi })^2\times 9.8\\\\L=0.993\ m[/tex]

So, the radio booth is 0.993 meters above the field.

The height of the radio booth is calculated to be approximately 9.9 cm by using the given oscillation period of 0.02 s and applying the pendulum formula.

A pendulum's period (T) is given by the formula:

T = 2π√(L/g)

Where:

Given that 1010 complete oscillations take 20.2 seconds, we can find the period:

Now, using the period formula, we rearrange to solve for L:

Square both sides to get:

Then solve for L:

Plugging in the values:

So, the height from which the microphone is hanging is approximately 9.9 cm.