the goose has a mass of 23.5 lb (pounds) and is flying at 11.5 miles/h (miles per hour). what is the kinetic energy of the goose in joules? enter your answer numerically in joules.

Answer:

7572 m/s

Explanation:

The force between two masses separated by a distance r is given as:

\(F=G\frac{m_1m_2}{r^2}\)

Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is \(6.674*10^-11 Nm^2/kg^2\)

The mass of the earth (\(m_1\)) is far greater than the mass of the sputnik (\(m_2\)). Therefore \(m_1m_2=m_1\). The mass of the sputnik is neglected, therefore:

\(F=G\frac{m_1}{r^2}=\frac{(6.674*10^{-11})(5.98*10^{24})}{(6.96*10^6)^2} = 8.2389N\)

But F is actually centripetal acceleration, a = v²/r

\(8.2389 = v^2 / 6.96*10^6\\v=7572m/s\)

The maximum force that can be exerted horizontally on the crate without moving it is 499.8 N, considering the static friction between the crate and the wood floor. Once the crate starts to slip, the magnitude of its acceleration will be 2.94 m/s^2 due to the kinetic friction between the crate and the wood floor.

(a) To determine the maximum force that can be exerted horizontally on the crate without moving it, we need to consider the static friction between the crate and the wood floor. The maximum force can be calculated using the formula:

F_max = μ_s * N,

where μ_s is the coefficient of static friction and N is the normal force acting on the crate.

The normal force N is equal to the weight of the crate, which can be calculated as:

N = m * g,

where m is the mass of the crate and g is the acceleration due to gravity.

Substituting the given values, we have:

N = 102 kg * 9.8 m/s^2 = 999.6 N.

Therefore, the maximum force that can be exerted horizontally on the crate without moving it is:

F_max = 0.5 * 999.6 N = 499.8 N.

(b) Once the crate starts to slip, the force of friction changes from static friction to kinetic friction. The magnitude of the crate's acceleration can be calculated using the formula:

a = (μ_k * g),

where μ_k is the coefficient of kinetic friction.

Substituting the given value, we have:

a = 0.3 * 9.8 m/s^2 = 2.94 m/s^2.

Therefore, if the applied force continues once the crate starts to slip, the magnitude of its acceleration will be 2.94 m/s^2.

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a. The probability that a randomly selected soda can has a volume less than 12 fluid ounces is approximately 0.3085

b. The probability that the sample mean volume is less than 12 fluid ounces, when a sample of size n = 32 is taken, is approximately 0.0023

c. The distribution of the sample mean becomes narrower and more concentrated around the population mean. Consequently, the probability of obtaining a sample mean less than 12 fluid ounces decreases because the sample mean is less likely to deviate significantly from the population mean.

a) Let X be the volume of a randomly selected soda can. We are given that the mean (μ) is 12.01 fluid ounces and the standard deviation (σ) is 0.02 fluid ounces.

We need to calculate P(X < 12). To do this, we standardize the variable using the z-score formula:

z = (X - μ) / σ

Substituting the given values, we have:

z = (12 - 12.01) / 0.02

= -0.5

Now, we can use a standard normal distribution table or calculator to find the probability associated with the z-score of -0.5. From the table, we find that the probability is approximately 0.3085.

b) When a sample of size n = 32 soda cans is drawn randomly from the population, the mean volume of the sample (denoted by X-bar) follows a normal distribution with the same mean (μ = 12.01 fluid ounces) but a smaller standard deviation (σ-bar) given by:

σ-bar = σ / sqrt(n)

Substituting the values, we have:

σ-bar = \(0.02 / \sqrt{(32)\)

= 0.02 / 5.6569

≈ 0.00354

Now, we need to calculate P(X-bar < 12). Again, we standardize the variable using the z-score formula:

z = (X-bar - μ) / σ-bar

Substituting the given values, we have:

z = (12 - 12.01) / 0.00354

≈ -2.8249

Using the standard normal distribution table or calculator, we find that the probability associated with the z-score of -2.8249 is approximately 0.0023.

c) As larger and larger sample sizes are taken, the probability that the sample mean volume is less than 12 fluid ounces tends to decrease. This is because as the sample size increases, the sample mean becomes a better estimate of the population mean. The larger the sample size, the more reliable and representative the sample mean is of the true mean. Hence, the sample mean is more likely to be closer to the population mean.

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Answer:

81.5

Explanation:

the total distance of 815km divided by total hours traveled 10hrs.

815 divided by 10 = 81.5

The kinetic energy is obtained as 1960 J.

The term energy refers to the ability to do work. There are several forms of energy. However, the mechanical energy refers to the energy that is either at rest or in motion.

Now we have the mechanical energy as the sum of the potential and kinetic energy. Thus;

Mechanical energy = 3200 J

Potential energy = 1260 J

Kinetic energy = 3200 J - 1260 J

= 1960 J

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Answer:

The answer is 10 m/s^2

Explanation:

Newtons Second Law is: F=ma

Taking into account the Newton's second law, the object’s  acceleration is 10 m/s².

Newton's Second Law, also called the Fundamental Law of Dynamics, establishes the relationship that exists between the forces that act on an object and the acceleration it experiences.

This law indicates that: "The change in motion is directly proportional to the printed motive force and occurs according to a straight line along which that force is printed."

This means that the acceleration of a moving object depends on the amount of force applied to it at a given moment.

Mathematically, Newton's second law is expressed by:

F= m×a

where:

In this case, you know:

Replacing:

10 N= 1 kg× a

Solving:

a= 10 N÷ 1 kg

a= 10 m/s²

Finally, the object’s  acceleration is 10 m/s².

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A constant force of 1.25N

Explanation:

This is a kinematics problem.

First find acceleration,

then use F = ma to find force.

Given:

mass = .5kg

delta x = 20m

t = 4s

a = ?

Friction = 0

From the kinematics equations:

delta x = Vi + (1/2)at^2

Plug in terms that are given:

20m = 0 + (1/2)a(4^2)

(2*20m)/(16s^2) = a

40m/(16s^2)= 2.5m/s^2

Now use F = ma to find force exerted on object.

F = (0.5kg)*(2.5m/s^2)

F = 1.25N

Answer:

direction

size

Explanation:

Vectors are physical quantities with both magnitude or size and direction.

Scalars are physical quantities with only size but not direction.

When describing a vector, on must specify its magnitude and direction.

Only the size of scalar quantities are needed to describe them.

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5200 × 25

We have the final answer as

Hope this helps you

3) 43

5) a - She should compare the increase in temperature...

(if a is wrong, try c)

The cT air mass would have the greater specific humidity; SPECIFIC HUMIDITY is the weight (amount) of vapour contained in a unit weight (amount) of air.

Relative humidity, which is frequently stated as a percentage, describes how humid it is right now in comparison to how humid it would be at its maximum temperature. The proportion of vapour mass to all moist air parcel mass is known as specific humidity. Surface longevity depends heavily on humidity.

The most accurate way to measure humidity is with a unit called the specific humidity unit. The specific humidity unit, g.kg-1, is used to quantify the weight of water vapor per unit weight of air. This measurement is made in grams of water vapor per kilogram of air.

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Hi There, Buddy!

To solve this problem, let us divide the entire motion into 2 phases

1.) When the elevator is moving with an acceleration

2.) When the elevator has a deceleration

For the first half -

u = 0 , a = 0.6 , t = t1

v = 0.6(t1)  ---------- ( 1 )

s = 1/2 x 0.6 x (t1)^2 = 0.3(t1^2) -------- ( 2 )

For the second half -

u = 0.6(t1) , a = -0.8 , v = 0

0.6 x (t1) = 0.8 (t2) ----> 3(t1) = 4(t2) ---------- ( 3 )

0.36(t1)^2 = 1.6 s2 ---------- ( 4 )

Now, we know that the total time taken is 10 seconds

------- t1 + t2 =10 , 3t1 = 4t2 ---- t1 = 40/7 , t2 = 30/7

Now, substitute values of t1 and t2 in s1 and s2 and add both s1 and s2 to get total distance travelled.

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Answer:

94 N-m

Explanation:

Joe is standing on the pedal of a bicycle. If his mass of 65 kg, the pedal makes an angle of 55º above the horizontal, and the pedal is 18 cm from the center of the chain ring. The amount of torque Joe exert is 94 N-m.

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is a torque.

Torque is defined as Γ = r×F = r.F.sin(θ). In other words, torque is the cross product of the force vector, where 'θ' is the angle between r and F, and the distance vector (the distance between the pivot point and the place where force is applied).

Given in question mass 65 Kg so force, F = mg = 637 N

Distance r = .18 m and  θ = 55 so sinθ = .82

Torque = rFsin(θ) putting the values, we get

Torque = 94 N-m.

The amount of torque Joe exert is 94 N-m.

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Answer:

Hi

Explanation:

A H20

Answer:

16.55 km

Explanation:

to find the displacement use the Pythagorean theorem to find the distance between the starting position and the ending position

15 becomes 225 when squared

7 is 49 when squared

225+49=274

sqrt of 274 is 16.55

The maximum possible speed of the train at the end of the first 4.0 km of the journey is 0 m/s.

To calculate the maximum possible speed of the train at the end of the first 4.0 km of the journey, we can apply the principle of conservation of energy.

Assuming there are no external forces like friction or air resistance, the initial potential energy of the train will be converted into kinetic energy.

The potential energy (PE) of the train at the beginning of the journey can be calculated as PE = mgh, where m is the mass of the train, g is the acceleration due to gravity (approximately \(9.8 m/s^2\)), and h is the height difference (in this case, we assume it to be zero).

The kinetic energy (KE) of the train at the end of the 4.0 km journey can be calculated as \(KE = (1/2)mv^2\), where v is the velocity of the train.

Since the potential energy is converted into kinetic energy, we can equate the two expressions:

PE = KE

\(mgh = (1/2)mv^2\)

Simplifying and canceling out the mass:

\(gh = (1/2)v^2\)

Substituting the values, \(g = 9.8 m/s^2\)and h = 0, we get:

\((9.8 m/s^2)(0) = (1/2)v^2\)

Simplifying further:

\(0 = (1/2)v^2\)

This equation tells us that the maximum possible speed of the train at the end of the first 4.0 km of the journey is 0 m/s.

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The uncertainty of energy of the gamma ray is 3.34 x 10⁻²⁵MeV.

Energy of the gamma ray emitted by the nucleus, E = 0.511 MeV

Lifetime of the nucleus, Δt = 1 ns = 10⁻⁹s

a) The expression for the uncertainty of energy of the gamma ray is given by,

ΔE = h/(2Δt)

ΔE = 6.67 x 10⁻³⁴/(2 x 10⁻⁹)

ΔE = 3.34 x 10⁻²⁵MeV

b) Detection of gamma rays is carried out photon by photon. By studying the impact they have on materials, gamma rays can be found.

Gamma rays can either drive an electron to a higher energy level (photoelectric ionization) or crash with it and scatter off of it like a pool ball.

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This is a multi-part question involving a three-dimensional harmonic oscillator and its angular momentum.

a. The orbital angular momentum operator `L` can be written in terms of the position and momentum operators as `L = r x p`. The squared magnitude of the angular momentum is given by `L^2 = Lx^2 + Ly^2 + Lz^2`. The `z` component of the angular momentum can be written as `Lz = xp_y - yp_x`, where `p_x` and `p_y` are the momentum operators along the `x` and `y` directions, respectively.

Since the state vector `|ψ⟩` is given as a product of quasi-classical states for one-dimensional harmonic oscillators along each axis, we can use the ladder operator formalism to evaluate the action of `Lz` on `|ψ⟩`. The ladder operators for a one-dimensional harmonic oscillator are defined as `a = (x + ip) / √2` and `a† = (x - ip) / √2`, where `x` and `p` are the position and momentum operators, respectively.

Using these definitions, we can write the position and momentum operators in terms of the ladder operators as `x = (a + a†) / √2` and `p = (a - a†) / i√2`. Substituting these expressions into the definition of `Lz`, we get:

`Lz = (xp_y - yp_x) = ((a_x + a_x†) / √2)((a_y - a_y†) / i√2) - ((a_y + a_y†) / √2)((a_x - a_x†) / i√2)`
  `  = (1/2i)(a_xa_y† - a_x†a_y - a_ya_x† + a_y†a_x)`
  `  = iħ(a_xa_y† - a_x†a_y)`

where we have used the commutation relation `[a, a†] = 1`.

The action of this operator on the state vector `|ψ⟩` is given by:

`(Lz)|ψ⟩ = iħ(a_xa_y† - a_x†a_y)|αx⟩|αy⟩|αz⟩`
        `= iħ(αxa_y† - αya_x†)|αx⟩|αy⟩|αz⟩`
        `= iħ(αx - αy)Lz|αx⟩|αy⟩|αz⟩`

where we have used the fact that the ladder operators act on quasi-classical states as `a|α⟩ = α|α⟩` and `a†|α⟩ = d/dα|α⟩`.

Since `(Lz)|ψ⟩ = iħ(αx - αy)Lz|ψ⟩`, it follows that `(Lz)^2|ψ⟩ = ħ^2(αx - αy)^2(Lz)^2|ψ⟩`. Therefore, we have:

`(L^2)|ψ⟩ = (Lx^2 + Ly^2 + Lz^2)|ψ⟩`
        `= ħ^2(αx^2 + αy^2 + (αx - αy)^2)(L^2)|ψ⟩`
        `= ħ^2(αx^2 + αy^2 + αx^2 - 2αxαy + αy^2)(L^2)|ψ⟩`
        `= ħ^2(3αx^2 + 3αy^2 - 4αxαy)(L^2)|ψ⟩`

This shows that `(L^2)|ψ⟩` is proportional to `(L^2)|ψ⟩`, which means that `(L^2)` is an eigenvalue of the operator `(L^2)` with eigenstate `|ψ⟩`. The eigenvalue is given by `(L^2) = ħ^2(3αx^2 + 3αy^2 - 4αxαy)`.

b. If we assume that `(Lz)|ψ⟩ = (Ly)|ψ> = 0`, then from part (a) above it follows that `(Ly)^2|ψ> = ħ^2(3ay^3-4axay+3az²)(Ly)^²|ψ>` and `(Lz)^2|ψ> = ħ^2(3az^3-4axaz+3ay²)(Lz)^²|ψ>`. Since `(Ly)^2|ψ> = (Lz)^2|ψ> = 0`, it follows that `3ay^3-4axay+3az² = 0` and `3az^3-4axaz+3ay² = 0`. Solving these equations simultaneously, we find that `ax = ay = az = 0`.

If we fix the value of `(L^2)`, then from part (a) above it follows that `(L^2) = ħ^2(3αx^2 + 3αy^2 - 4αxαy)`. Since `ax = ay = az = 0`, this equation reduces to `(L^2) = 0`.

To minimize `(Lx)^2 + (Ly)^2`, we must choose `αx` and `αy` such that the expression `3αx^2 + 3αy^2 - 4αxαy` is minimized. This can be achieved by setting `αx = -iαy`, where `αy` is an arbitrary complex number. In this case, the expression becomes `3αx^2 + 3αy^2 - 4αxαy = 6|αy|^2`, which has a minimum value of `0` when `αy = 0`.

c. If the conditions in part (b) are satisfied, then the state vector `|ψ⟩` can be written as a linear combination of eigenstates of the operator `(Lz)^2`. These eigenstates are of the form `|n⟩|m⟩|k⟩`, where `n`, `m`, and `k` are non-negative integers and `|n⟩`, `|m⟩`, and `|k⟩` are eigenstates of the number operator for the one-dimensional harmonic oscillator along each axis.

The action of the ladder operators on these states is given by:

`a_x|n⟩|m⟩|k⟩ = √n|n-1⟩|m⟩|k⟩`
`a_x†|n⟩|m⟩|k⟩ = √(n+1)|n+1⟩|m⟩|k⟩`
`a_y|n⟩|m⟩|k⟩ = √m|n⟩|m-1⟩|k⟩`
`a_y†|n⟩|m⟩|k⟩ = √(m+1)|n⟩|m+1⟩|k⟩`
`a_z|n⟩|m⟩|k⟩ = √k|n⟩|m⟩|k-1⟩`
`a_z†|n⟩|m⟩|k⟩ = √(k+1)|n⟩|m⟩|k+1⟩`

Since we have assumed that `(Lz)|ψ> = (Ly)|ψ> = 0`, it follows that:

`(Lz)|ψ> = iħ(a_xa_y† - a_x†a_y)|ψ> = iħ(∑_n,m,k c_nmk(a_xa_y† - a_x†a_y)|n>|m>|k>)`
        `= iħ(∑_n,m,k c_nmk(√n√(m+1)|n-1>|m+1>|k> - √(n+1)√m |n+1>|m-1>|k>))`
        `= iħ(∑_n,m,k (c_(n+1)(m+1)k√(n+1)√(m+1) - c_n(m-1)k√n√m)|n>|m>|k>)`
        `= 0`

This implies that for all values of `n`, `m`, and `k`, we must have:

`c_(n+1)(m+1)k√(n+1)√(m+1) - c_n(m-1)k√n√m = 0`

Similarly, since `(Ly)|ψ> = 0`, it follows that:

`(Ly)|ψ> = iħ(a_xa_z† - a_x†a_z)|ψ> = iħ(∑_n,m,k c_nmk(a_xa_z† - a_x†a_z)|n>|m>|k>)`
        `= iħ(∑_n,m,k c_nmk(√n√(k+1)|n-1>|m>|k+1> - √(n+1)

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The plane traveled 2125 km north and 1812 km west after 3.40 hours.

To solve this problem, we need to break down the velocity vector into its northerly and westerly components. We can use trigonometry to do this. Let's call the northerly component "v north" and the westerly component "v west". We can find v north by multiplying the velocity (835 km/h) by the sine of the angle (41.5∘) between the velocity vector and the north direction. Similarly, we can find v west by multiplying the velocity by the cosine of the angle.

Using the trigonometric functions, we get:
v north = 835 km/h x sin(41.5∘) = 625 km/h
v west = 835 km/h x cos(41.5∘) = 533 km/h

Now, to find how far north and how far west the plane has traveled after 3.40 hours, we simply need to multiply the components by the time:
Δd north = v north x time = 625 km/h x 3.40 h = 2125 km
Δd west = v west x time = 533 km/h x 3.40 h = 1812 km

So the plane traveled 2125 km north and 1812 km west after 3.40 hours.

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Starting from rest on a circular track with a radius of 300 m and a constant tangential acceleration of 0.75 m/s², the car will travel a distance of approximately 0.2119 meters or 21.19 centimeters in 0.75 seconds.

To determine the distance traveled and the time elapsed by the car starting from rest on a circular track with a radius of 300 m and a constant tangential acceleration of 0.75 m/s², we can use the equations of circular motion.

The tangential acceleration is the rate of change of tangential velocity. Since the car starts from rest, its initial tangential velocity is zero (v₀ = 0).

Using the equation:

v = v₀ + at

where v is the final tangential velocity, v₀ is the initial tangential velocity, a is the tangential acceleration, and t is the time, we can solve for v:

v = 0 + (0.75 m/s²) * t

v = 0.75t m/s

The tangential velocity is related to the angular velocity (ω) and the radius (r) of the circular track:

v = ωr

Substituting the values:

0.75t = ω * 300

Since the car starts from rest, the initial angular velocity (ω₀) is zero. So, we have:

ω = ω₀ + αt

ω = 0 + (0.75 m/s²) * t

ω = 0.75t rad/s

We can now substitute the value of ω into the equation:

0.75t = (0.75t) * 300

Simplifying the equation gives:

0.75t = 225t

t = 0.75 seconds

The time elapsed is 0.75 seconds.

To calculate the distance traveled (s), we can use the equation:

s = v₀t + (1/2)at²

Since the initial velocity (v₀) is zero, the equation becomes:

s = (1/2)at²

s = (1/2)(0.75 m/s²)(0.75 s)²

s = (1/2)(0.75 m/s²)(0.5625 s²)

s = 0.2119 meters or approximately 21.19 centimeters

Therefore, the car travels a distance of approximately 0.2119 meters or 21.19 centimeters.

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Answer:

F = I L B     describes the perpendicular force on a wire of length L in a uniform field B

B = F / (I L) = 1.2 / (9 * 2.5) = .053 Tesla

The electric force between two charges can be calculated using Coulomb's law, which states that the force (F) between two charges (q1 and q2) separated by a distance (r) is given by:

F = (k * |q1 * q2|) /\(r^2\)

where k is the Coulomb constant.

In this case, the charge on the balloon is -8.0 µC and the charge on the denim is assumed to be positive, so q1 = -8.0 µC and q2 = +8.0 µC (equal in magnitude). The distance between the charges is 4.0 cm, or 0.04 m.

Plugging these values into Coulomb's law:

F = (8.98755 × 10^9 N · \(m^2/C^2\) * |(-8.0 µC) * (+8.0 µC)|) / (0.04 \(m)^2\)

Simplifying the expression inside the parentheses:

|(-8.0 µC) * (+8.0 µC)| = (8.0 µ\(C)^2\) = (8.0 × \(10^-^6 C)^2\) = 6.4 × \(10^-^1^1 C^2\)

Now substituting this value into the equation:

F = (8.98755 × \(10^9\) N · \(m^2/C^2\) * 6.4 × \(10^-^1^1 C^2\) / (0.04 \(m)^2\)

Calculating the numerator:

8.98755 × \(10^9\)N · \(m^2/C^2\)

* 6.4 × \(10^-^1^1 C^2\) = 5.751872 N · \(m^2\)

Calculating the denominator:

(0.04 \(m)^2\)= 0.0016\(m^2\)

Now dividing the numerator by the denominator:

F = 5.751872 N · \(m^2\)/ 0.0016\(m^2\)≈ 3594.92 N

Therefore, the electric force between the balloon and the denim, when separated by a distance of 4.0 cm, is approximately 3594.92 N. The force is attractive since the charges have opposite signs (negative and positive).

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Answer:

What is the frequency of ocean waves that have a speed of 18 m/s and wavelength of 50 m/s?

0.36hz

The frequency of the wave is 0.36 Hz.

Frequency of a wave is defined as the number of oscillations completed by the wave in one second.

Here,

Speed of the wave, v = 18 m/s

Wavelength of the wave, λ = 50 m

We know that, the equation for the speed of a wave is given as,

v = fλ

where f is the frequency of the wave.

Therefore, frequency,

f = v/λ

f = 18/50

f = 0.36 Hz

Hence,

The frequency of the wave is 0.36 Hz.

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Increase the frequency of the incident light. The photoelectric effect occurs when light of a sufficient frequency strikes a metal and frees electrons from the metal's surface. The energy of the incident photons must be greater than the energy required to remove the electrons, known as the work function. Thus, increasing the frequency of the incident light will result in higher energy photons, increasing the likelihood of electron ejection in the photoelectric effect.

The kinetic energy of the photoelectrons that are released will grow as the light amplitude increases, according to the wave model of light, yet the measured current will increase as the frequency increases. In contrast to expectations, studies revealed that raising the light's frequency and amplitude raised the current and the photoelectrons' kinetic energy, respectively.

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Answer:

Describe the limitations in detail in their final report

Explanation:

just did test

Answer:

The field will remain the same

Explanation:

This is because electric field given as

E1= kq1/r²

And that of second charge

E² = kq2/r²

Is not affected by the size of the second charge q2

Squeezing the clutch lever to slow down is best to be done when approaching a curve and is denoted as option C.

This is a mechanical device which used in the engagement and disengagement of transmission system in a vehicle.

Slowing down will give the driver more time to maneuver the curves so as to prevent accident. This is why it is imperative to squeeze the clutch in such road condition.

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Answer:

c

Explanation:

mark brainliest of your straight

Answer:

the answer is D

Explanation: