How many significant figures does the number 0.0030 have?

A) 1
B) 2
C) 4
D) 3

The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.

To find the change in momentum for the baseball, we can use the formula:

Change in momentum = Final momentum - Initial momentum

Momentum is defined as the product of mass and velocity.

Given data:

Mass of the baseball (m) = 0.140 kg

Initial velocity of the baseball (\(v_i_n_i_t_i_a_l)\) = 30.0 m/s

Final velocity of the baseball (\(v_f_i_n_a_l_\)) = -44.0 m/s (negative sign indicates opposite direction)

To calculate the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = m * \(v_i_n_i_t_i_a_l\) = 0.140 kg * 30.0 m/s = 4.2 kg·m/s

To calculate the final momentum, we multiply the mass by the final velocity:

Final momentum = m * \(v_f_i_n_a_l_\) = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s

Now we can find the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)

Change in momentum = -10.36 kg·m/s

Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.

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Answer:

it is double the temperature change of iron

For the graph of Annual cray fish caught in Lambert Bay for the last 20 years, the axes are as follows:

Y-axis: amount of fish caught

X-axis: year

A graph is a pictorial illustration of data.

There are different types of graphs such as

A graph has the vertical axis, known as the Y-axis, and the horizontal axis known as the X-axis.

For the graph of Annual cray fish caught in Lambert Bay for the last 20 years, the axis can be labelled as follows:

Y-axis: amount of fish caught

X-axis : year

In conclusion, in a graph, the vertical axis is known as the Y-axis while  the horizontal axis is known as the X-axis.

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The acceleration of the dragster is 2.01 * 10^5 m/s^2

The term acceleration refers to the change in velocity with time. Hence the formula for acceleration is given as;

a = v - u/t

a = acceleration

v = final velocity

u = initial velocity

t = time taken

Now;

v = 100 miles or 160934 meters

u = 0 miles or 0 meters

t = 0.8 seconds

a =  160934 - 0/ 0.8

a = 2.01 * 10^5 m/s^2

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Answer:

630dB

Explanation:

First you must know that the relationship between the sound intensity and the distance is an inverse relationship.

Let I represent the sound intensity

d represent the distance

Since I is inversely proportional to d, hence;

I ∝ 1/d

I = k/d

k is the constant of proportionality

If the whistle has a sound intensity of 90 dB when the boat passes you 7 m away, this means;

when I = 90dB, d = 7m

Substitute and get k first;

I = k/d

90 = k/d

k = 90 * 7

k = 630

To get how loud (in dB) is it in the cockpit just 1 m from the whistle, we will find I when d = 1 m

From the expression I = k/d

k = 630, d = 1

I = 630/1

I = 630dB

Hence the intensity is 630bB loud in the cockpit when it is just 1m from the whistle.

Answer:

africa

Explanation:

África África ooo

Answer:

100

w=f*s

20*5=100....

100 Joules of work is done on the cart.

Work done by a force is defined as the product of the displacement and the component of the applied force on the object in the direction of displacement. When we push a block with some force, the body moves with some acceleration, so it is called work done.

Work done is expressed as W=Fd and its unit is joules which can be defined as the amount of work done by a force in Newton is applied to an object, as a result of which it is displaced in meter.

For above given information,

Force= 20 N

Distance= 5 m

So, work done= 20*5= 100 Joules

Thus, 100 Joules of work is done on the cart.

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Explanation:

The triple beam balance is used to measure masses very precisely; the reading error is 0.05g

It as dangerous to be hit by the gun as by the bullet because of the following;

(A) The butt distributes the recoil force over an area much larger than that of the bullet.

(B) The rifle has a much lower speed than the bullet.

The principle of conservation of linear momentum states that in an isolated system, the total momentum of the system is conserved.

That is the sum of the initial momentum is equal to the sum of the final momentum.

momentum of the gun = momentum of the bullet

Mu = mU

where;

If the forward momentum of a bullet is the same as the backward momentum of the gun, the speed of the gun will be smaller than the speed of the bullet since the mass of the gun is bigger than mass of the bullet.

We cannot conclude on the kinetic energy, since it depends on both mass and velocity.

Finally, the butt distributes the recoil force over an area much larger than that of the bullet, since the butt has a larger surface area and will hit more surface area than the bullet.

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In Tesla squared meters, the magnitude of the change in the magnetic flux. (ΔΦ= 2.3468T⋅m²).

The magnetic flux through the loop is given by;

Φ = B A cosθ

where B is magnetic field strength, A is area of the loop, and θ is angle between the normal to the loop and the direction of the magnetic field. Since the magnetic field is perpendicular to the plane of the loop, θ = 0 and cosθ = 1.

The magnitude of the change in the magnetic flux is given by;

ΔΦ = Φ₂ - Φ₁

where Φ₁ is the initial magnetic flux through the loop when the magnetic field strength is B₁ = 0.65 T, and Φ₂ is the final magnetic flux through the loop when the magnetic field strength is B₂ = 4.5 T.

The initial magnetic flux through the loop is;

Φ₁ = B₁ A cosθ = 0.65 × πr² × 1 = 0.1126 T⋅m²

The final magnetic flux through the loop is;

Φ₂ = B₂ A cosθ = 4.5 × πr² × 1 = 2.4594 T⋅m²

Therefore, the magnitude of the change in the magnetic flux is;

ΔΦ = Φ₂ - Φ₁ = 2.4594 - 0.1126 = 2.3468 T⋅m²

Therefore, ΔΦ = 2.3468 T⋅m².

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Answer:

Explanation:

read the assignment and use this

Topic sentence

Evidence from the text

Thoughts in your own words

Opinion

Wrap up sentence

T

E

T

O

W

and 2 of those make 2 paragraphs

Answer:

The phase existing as small droplets is called the dispersed phase and the surrounding liquid is known as the continuous phase. Emulsions are commonly classified as oil-in-water (O/W) or water-in-oil (W/O) depending on whether the continuous phase is water or oil.

Explanation:

A collision that is elastic occurs when there is no net loss of kinetic energy in the system as a result of the collision. Kinetic energy and momentum are both conserved in elastic collisions.

When two balls collide at a pool table, that is an instance of an elastic collision. When you throw a ball on the ground and it bounces back into your hand, there is no net change in the kinetic energy, making it an elastic collision.

Two balls colliding at a pool table is an example of an elastic collision. When a ball is tossed to the ground and subsequently returns to your hand, there is no net change in the kinetic energy, making it an elastic collision.

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Answer: Velocity is a vector; the word decreasing only applies to its magnitude (not its direction) which is called speed. If speed is increasing and direction is not changing, then acceleration is positive. If speed is constant then acceleration is zero. If speed is decreasing then acceleration is negative.

Explanation:

Answer:

W =  343.2 J

Explanation:

Given that,

Mass of bale of hay = 26 kg

Horizontal force exerted = 88 N

Distance moved, d = 3.9 m

Work done, W = Fd

Put all the values,

W = 88 N × 3.9 m

= 343.2 J

So, the work done is 343.2 J.  

Given

2 identical charged particles get closer to each other

To find

The strength of the electrical force between them.

Explanation

The electrostatic force is indirectly proportional to the square of the distance between the charges.

So as the distance decreases, the electric force increases.

Conclusion

The correct opttion is

A. increases

Answer:

The word is "wind".

Explanation:

The sentence would be:

"The sun is also the reason wind exists. The wind is moving air."

Hope this helped !

Answer:

C. the help our understanding of the formation of planetary bodies.

Explanation:

this is the answer of PLATO

A transverse wave moves perpendicular to the particle vibration, while a compressional wave moves parallel to the particle vibration. Light waves are an example of a transverse wave, and sound waves are an example of a compressional wave.

A transverse wave is a type of wave in which the particles of the medium vibrate perpendicular to the direction of the wave's propagation. This means that as the wave moves forward, the particles move up and down or side to side. An example of a transverse wave is a light wave. When light travels through space, the electric and magnetic fields oscillate perpendicular to the direction of the wave.
A compressional wave is a type of wave in which the particles of the medium vibrate parallel to the direction of the wave's propagation. This means that as the wave moves forward, the particles move back and forth, creating compressions and rarefactions in the medium. An example of a compressional wave is a sound wave. When sound travels through air, the molecules of air vibrate back and forth, creating areas of high pressure (compressions) and low pressure (rarefactions).
In summary, a transverse wave moves perpendicular to the particle vibration, while a compressional wave moves parallel to the particle vibration. Light waves are an example of a transverse wave, and sound waves are an example of a compressional wave.

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Answer: λ (wavelength) = 3.3 m →  ΔS = 3.3 m

v (speed) = 5.6 m/s → ΔV = 5.6 m/s

T (period) → ΔT = ?

f (frequency) = ?

If:

Now: (if, ΔT = T), Using the formula of the period of a wave, find the frequency:

Explanation: Your Welcome u.u

The frequency of this wave is approximately 1.697 Hz.

To find the frequency of a wave, we can use the formula:

frequency (f) = speed (v) / wavelength (λ)

Given:

Wavelength (λ) = 3.3 m

Speed (v) = 5.6 m/s

Substituting these values into the formula, we can calculate the frequency:

f = v / λ

f = 5.6 m/s / 3.3 m

f ≈ 1.697 Hz

Therefore, the frequency of this wave is approximately 1.697 Hz.

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Answer:

8 + 6 = 14 miles north

Explanation:

a) To find the net charge on the sphere, we first need to find the total area of the surface. The surface area of a sphere is given by:

A = 4πr^2

where r is the radius of the sphere. In this case, the sphere has a diameter of 1.2 m, so the radius is 0.6 m. Therefore, the surface area of the sphere is:

A = 4π(0.6 m)^2 = 4.52 m^2

The total charge on the surface can be found by multiplying the charge density by the surface area:

Q = σA = (8.1 µC/m^2) (4.52 m^2) = 36.6 µC

Therefore, the net charge on the sphere is 36.6 µC.

b) The electric flux emanating from the surface of the sphere can be found using Gauss's Law, which states that the electric flux through any closed surface is proportional to the charge enclosed within the surface.

Since the sphere has a uniform surface charge density, we can assume that the charge is evenly distributed throughout the entire surface. Therefore, we can imagine a hypothetical sphere of radius r < 0.6 m concentric with the original sphere, and use this to find the electric flux through the surface.

The electric field at a distance r from the center of a uniformly charged sphere is given by:

E = kQr / R^3

where k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the total charge on the sphere, and R is the radius of the sphere. Using the values Q = 36.6 µC and R = 0.6 m, we can solve for the electric field at a radius r:

E = (8.99 x 10^9 Nm^2/C^2)(36.6 x 10^-6 C) / (0.6 m)^3 * r^2

The electric flux through a spherical surface of radius r is given by:

Φ = E A = E 4π r^2

Substituting the expression for E, we get:

Φ = (8.99 x 10^9 Nm^2/C^2)(36.6 x 10^-6 C) / (0.6 m)^3 * 4π r^2

Simplifying and cancelling terms, we get:

Φ = 4.27 x 10^5 Nm^2/C

Therefore, the total electric flux emanating from the surface of the sphere is 4.27 x 10^5 Nm^2/C.

The net force on q₂ will be  1.07 x 10⁻² N, pointing to the left.

To find the net force on particle q₂, we need to calculate the force due to q₁  and q₃ individually and then add them up vectorially. We can use Coulomb's law to calculate the force between two point charges:

F = k × (q₁ × q₂) / r²

where F is the magnitude of the force, k is Coulomb's constant (k = 8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the two particles, and r is the distance between them.

The force due to q₁ on q₂ can be calculated as:

F₁ = k × (q₁ × q₂) / r₁²

where r1 is the distance between q₁ and q₂ (r₁ = 0.180 m).

Similarly, the force due to q₃ on q₂ can be calculated as:

F₂ = k × (q₃ × q₂) / r₃²

where r₃ is the distance between q₂ and q₃ (r₃= 0.230 m).

The direction of each force can be determined by the direction of the electric field due to each charge. Since q₁ and q₃ have opposite signs, their electric fields point in opposite directions. Therefore, the force due to q₁ points to the left and the force due to q₃ points to the right.

To find the net force, we need to add up the forces vectorially. Since the forces due to q₁ and q₃ are in opposite directions, we can subtract the magnitude of the force due to q₃ from the magnitude of the force due to q₁ to get the net force on q₂:

Fnet = F₁ - F₃

Substituting the values we get:

Fnet = k × (q₁ × q₂) / r₁² - k × (q₃ × q₂) / r₃²

Plugging in the values we get:

Fnet = (8.99 x 10⁹ Nm²/C²) × [(9.33 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.180 m)² - (-8.42 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.230 m)²]

Fnet = 1.07 x 10⁻² N

Therefore, the net force on q₂ is 1.07 x 10⁻² N, pointing to the left.

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Answer:

F=w=ma                                                                                                                                                       OR by using equations of motions                                                                               vf=vi-at                    : a=vf-vi/t                                                              eq 1                         s=vit+1/2at squre                                                                                 eq 2                     2as=vf squre - vi squre                                                                        eq 3                                                                                                                                                                                  

Explanation:

where m is the mass of falling body , f is the weight is the force acting down ward , vf is the final velocity, vi is the inetial velocity , t is the time and s is the distance covered by a body.

Answer:

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA

Initial diameter of the wire, d₁ = 4 mm = 0.004 m

Final diameter of the wire, d₂ = 1 mm = 0.001 m

Length of wire, L = 2.00 m

Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

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Explanation:

The changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

As the electrons traverse one end of the wire to another, their mean drift velocity undergoes a shift of 3.506 x 10⁻⁷ m/s with an average acceleration of 4.38 x 10⁻¹⁵ m/s² in accordance with the following parameters:

- The current flowing through the wire is at 4.00 mA.

- The original diameter of the wire, d₁, measures at 4 mm or 0.004 m.

- Conversely, the final diameter, d₂, displays a measurement of 1 mm or 0.001 m.

- The length of the entire wire is consistent, measuring at 2 meters.

- Notably, the density of electrons present within copper reaches an estimated value of 8.5 x 10²⁸ /m³.

Calculations regarding both initial and final area coverage provided by copper must be explored along with numerical data involving the two varying drift velocities for accurate results.

Thus, we arrive at the change rate of the mean drift velocity between points in the wire as well as the plenitude of electron acceleration achieved after contemplation into the corresponding motion periods.

The conclusion reflects that our measurements find the changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

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The material that obeys Hooke's law is elastic material.

Hooke's Law says that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance. And it is known for the formula  F = kX

F is the force, x is the deformation, and k is the spring constant.

Elastic materials are able to return to their original shape after being deformed by a force.

The amount of deformation is noted to be directly proportional to the amount of force applied.

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Answer:

214.72m

Explanation:

The range (R) of a projectile can be calculated using the formula:

R = u²sin2θ/g

Where R = Horizontal range

u = initial velocity

θ = angle of initial velocity

g = acceleration due to gravity (9.8m/s²)

According to the provided information in this question, R= ?, u= 51m/s, θ = 63°, g = 9.8m/s².

Hence, R = u²sin2θ/g

R = 51² × sin 2×63/ 9.8

R = 2601 × sin 126/9.8

R = 2601 × 0.809/9.8

R = 2104.25/9.8

R = 214.72m

The mass of the wire of lowest A on a piano is 0.00165 kg.

The frequency of a vibrating string is given by the equation:

f = (1/2L) * sqrt(T/μ)

where f is the frequency of the string, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string (mass per unit length).

We know the frequency of the lowest A on a piano is 27.5 Hz. We also know that one half wavelength occupies the string, so the length of the string is half the wavelength:

L = (1/2) * λ

The wavelength of a sound wave is given by:

λ = 2L/n

where n is the number of nodes (points of zero displacement) in the wave. For the lowest A on a piano, n = 1, so we can write:

λ = 2L

Substituting this into the equation above for L, we obtain:

L = λ/2

Now we can substitute these values into the first equation:

27.5 = (1/2)(λ/2) * sqrt(308/μ)

Simplifying, we get:

λ = 4L

308/μ = 4(27.5)^2 (1/4)

μ = 0.000824 kg/m.

Since μ = m/L, where m is the mass of the wire and L is its length, we can find the mass of the wire by multiplying the linear mass density by the length of the string:

m = μL

The length of the string is given as 2.00 m, so we can write:

m = 0.000824 kg/m * 2.00 m = 0.00165 kg

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Answer:

flat lol, that's the answer