What net external force is exerted on a 1350-kg artillery shell fired from a battleship if the shell is accelerated at 2.00 ✕ 104 m/s2? (Enter the magnitude.)
N

What is the magnitude of the force exerted on the ship by the artillery shell?
N

Answer:

(B)  (the leaves will separate when subjected to a net charge)

The image that represents a gaining of a charge on the leaves of the electroscope by conduction is image B.

An electroscope is a device that detects the presence of an electric charge or ionizing radiation by suspending a pair of thin gold leaves from an electrical conductor that leads to the outside of an insulating container.

When an electrical charge is brought close to or in contact with a conductor, the leaves in an electroscope spread apart.

This occurs because the same charge, whether positive or negative, is received by the leaves, causing their repulsion (spreading apart).

A negatively charged rod is brought in close proximity to an uncharged electroscope. As a result, the electroscope leaves become further apart.

Thus, image B is representing the given scenario.

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An object must move in the same direction as the force is applied for work to be accomplished. Option C is correct.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative. it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.

The work done is found as;

W= F × d

Where.

F is the force

d is the displacement

W is the work done

In order for work to be done, an object must move in the same direction, and the force is applied

Hence option C is correct.

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Answer:

[tex]\huge\boxed{\sf a = 0.7 \ m/s^2}[/tex]

Explanation:

Initial velocity = [tex]V_i[/tex] = 0.5 m/s

Final velocity = [tex]V_f[/tex] = 4 m/s

Time = t = 5 s

Acceleration = a = ?

[tex]\displaystyle a =\frac{V_f-V_i}{t}[/tex]

Put the givens in the formula.

[tex]\displaystyle a=\frac{4-0.5}{5} \\\\a = \frac{3.5}{5} \\\\a = 0.7 \ m/s^2\\\\\rule[225]{225}{2}[/tex]  

Answer:

Sparks often indicate the presence of a high voltage, or "potential field". The higher the voltage; the farther a spark can jump across a gap, and with enough energy supplied can lead to greater discharges such as a glow or an arc.

Explanation:

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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The formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.

The impulse exerted on the ball at any given instant is calculated as follows;

J = ΔP

J = Pf - Pi

where;

Thus, the formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.

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Answer:

Power = 2000J/s

Explanation:

Power= Total Work done/Time

Given

mass = 100kg

distance(height) = 10m

time = 5s

acceleration due to gravity = 10m/s²

Work done = force × distance

Finding Force

Force = mass × acceleration due to gravity

Force = 100kg × 10m/s²

Force = 1,000N

Finding work done

Work done = Force × distance

Work done = 1,000N × 10m

Work done = 10,000J

Finding Power

Power = work done/time

Power = 10,000J/5s

Power = 2,000J/s

Therefore Power is 2,000J/s

Answer:kinetic energy

Explanation:

The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.

The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.

Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.

The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.

Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.

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Answer:

STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.

Explanation:

Explanation:

Answer:

antibiotics

Explanation:

im not sure if this is right hope this helps lol

Answer:

The value of work being done on the object is 958J.

Explanation:

Work done is equal to force multiply by distance, but when the angle is between the force and distance work done=Force (cos theta) × distance

Work done = Force(cos theta) × distance

Work done = 25N(cos 40.0°) × 50m

Work done = 25N(0.7660) × 50m

Work done = 19.15N × 50m

Work done = 957.5J = 958J

Therefore the value of work being done on the object is 958J.

After time 1.74 seconds ball will strike the ground.

A ball is thrown directly downward and information we have-

Initial velocity of ball (u) = 8.75m/s

distance/height (s) = 29.4m

Ball is thrown downward means there are gravitation force also worked and the acceleration of ball is-

acceleration (a) = g = 9.8 m/s²

When ball strike the ground then it's velocity should be 0 means-

Final velocity of ball (v) = 0 m/s²

For getting time period we can use 2nd equation of motion as-

s = ut + (1/2)at²

Now, 29.4 = (8.75 × t) + (1/2× 9.8 × t²)

29.4 = 8.75t + 4.9t²

Now we have a quadratic equation as -

4.9t² + 8.75t - 29.4 = 0

After solving it we get two values of t -

time = 1.74 and time = -3.5

But time can not be negative so we will reject the negative values.

so t = 1.74 seconds

So to conclude that after applying the second equation of motion the time taken by the ball to reach ground is 1.74 seconds.

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The area of the rectangle obtained is (3.75 ± 0.03) m².  The perimeter of the rectangle obtained is (8 ± 0.64) m.

Given dimensions:

Length (L) = (2.5 ± 0.2) m

Width (W) = (1.5 ± 0.2) m

(a) Area of the rectangle (A) is given by:

A = L × W

A = (2.5 ) × (1.5 )

A = 3.75 m²

The relative uncertainty in area = ΔA / A

ΔA = (ΔL / L) × A + (ΔW / W) × A

ΔA = (0.2 / 2.5) × 3.75 + (0.2 / 1.5) × 3.75

ΔA = 0.03 m²

Therefore, the area of the rectangle is (3.75 ± 0.03) m².

(b) Perimeter of the rectangle (P) is given by:

P = 2 × (L + W)

P = 2 × (2.5 + 1.5)

P = 2 × 4

P = 8 m

Relative uncertainty in perimeter = ΔP / P

ΔP = (ΔL / L) × P + (ΔW / W) × P

ΔP = (0.2 / 2.5) × 8+ (0.2 / 1.5) × 8

ΔP = 0.64 m

Therefore, the perimeter of the rectangle is (8 ± 0.64) m.

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Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that

[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]

Solve for [tex]a[/tex] :

[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]

The mass of block 1 is 1.30176 kg.

A tension force in physics is a force developed in a rope, string, or cable when stretched under an applied force.

Tension is acted along the length of the rope/cable in a direction that is opposite to the force applied on it.

According to the question,

Mass of block 2= 11.3 kg

Kinetic friction μ=0.200

The component of the weight in the direction of the incline

W = m₂gsinθ

T = [tex]F_{f} - W[/tex]........(1)

When the force of friction

[tex]F_{f}[/tex] = μ[tex]_k[/tex]N

[tex]F_{f}[/tex] = μ[tex]_k[/tex](m₂gcosθ)

Here , Put the value in equation 1

T = m₂gcos 31.5° - μ[tex]_k[/tex]m₂gcos 31.5°

We know that ,

T - m₁g = 0

T = m₁g

So,

m₁g = m₂g (Sin 31.5° - 0.200 x cos 31.5°)

m₁ = 11.3 x (0.0840 - 0.2 x 0.996)

m₁ = 11.3 x (0.1152)

m₁ = 1.30176 kg

Hence, the mass of block 1 is 1.30176 kg.

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An externality, sometimes called a spillover, occurs when an exchange between a buyer and seller has an impact on a third party who is not part of the exchange. Externalities can be positive or negative

Externalities refers to situations when the effect of production or consumption of goods and services imposes costs or benefits on others which are not reflected in the prices charged for the goods and services being provided.

Externalities Private markets offer an efficient way to put buyers and sellers together and determine what goods are produced, how they are produced, and who gets them. The principle that voluntary exchange benefits both buyers and sellers is a fundamental building block of the economic way of thinking. But what happens when a voluntary exchange affects a third party who is neither the buyer nor the seller?

Externalities can be negative or positive. If you hate country music, then having it waft into your house every night would be a negative externality. If you love country music, then what amounts to a series of free concerts would be a positive externality.

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Answer:

Gravitational potential energy is greater for the rock [tex]20\; {\rm m}[/tex] above the river.

Explanation:

The gravitational field near the surface of the earth is approximately uniform. Let [tex]g[/tex] denote the strength of the gravitational field near the surface of the earth.

Let [tex]m[/tex] denote the mass of each rock. Assume that the gravitational potential energy of the two rocks is [tex]0[/tex] when the rocks are at the surface of the river.

When the rock is at a height of [tex]h[/tex] above the river, the gravitational potential energy ([tex]\text{GPE}[/tex]) of that rock would be [tex]\text{GPE} = m\, g\, h[/tex]. In other words, the [tex]\text{GPE}\![/tex] of the rock is proportional to its height relative to the ground.

Hence, among the two rocks, [tex]\text{GPE}[/tex] would be greater for the rock that is higher above the ground ([tex]20\; {\rm m}[/tex]) than the rock [tex]10\; {\rm m}[/tex] above the ground.

a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω

b)ω = [tex]\frac{3g}{2l}[/tex]

c)14.7  the magnitude of the initial acceleration of the end B when the string is cut.

Acceleration is the rate of change of the velocity of an object with respect to time.

Given

A rod of length 56.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 cm long, one at each end of the rod.

When the string is cut rod rotates about other end.

Let's take momentum equation about that end:

mg*[tex]\frac{1}{2}[/tex] = Iω [I = [tex]ml^\frac{2}{3}[/tex]]

a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω

Angular acceleration:

b)ω = [tex]\frac{3g}{2l}[/tex]

c)Acceleration of end B:

a = lω= [tex]\frac{3g}{2}[/tex] =[tex]\frac{3*9.8}{2}[/tex]

a = 14.7 [g=9.8]

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The speed of the particle when it enters the magnetic-field region is 120,000 m/s.

Magnetic force on the particle is given as;

F = qvB

where;

From Newton's second law, force on an object is given as;

F = ma

where;

ma = qvB

v = ma/qB

v = (3 x 10⁻² x 12)/(3 x 10⁻⁶ x 1)

v = 120,000 m/s

Thus, the speed of the particle when it enters the magnetic-field region is 120,000 m/s.

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The total displacement for the trip is 5510 m

During part 1, the distance travelled is

(1/2) a t^2

= (1/2)(2.77)(225)

= 311.6 m

The final speed gotten is (2.77)(15)

= 41.55 m/s.

The average speed for that interval is half of that or 20.8 m/s

During part 2, the distance traveled is

41.55 m/s*2.05 min*60 sec/min

= 5110 m/s

During part 3, the speed drops from 41.55 m/s to

41.55 - (9.47)(4.39)

= 0

The average speed during this interval is

41.55/2 - 20.78 m/s

The distance travelled is 20.78*4.39

= 91.2 m/s

Total distance travelled = 312 + 5110 + 91 = 5513 m

The average speeds for legs 1,2,and 3 of the trip as well as for the complete trip are:

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Answer:

True

Explanation:

it is true because blabla bla

The reference point that should be used to describe Melanie motion is the pond (option C).

Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.

According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.

However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.

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The x-component for the vector given in the image is : [tex]B_{x}[/tex] = 18.3m

A Vector can be defined as a quantity that possesses both magnitude and direction thereby giving more details.

A vector is very important when it comes to dealing with forces.

Analysis

[tex]B_{x}[/tex] = B cos(Θ)

Θ = 180° - 170° = 10°

[tex]B_{x}[/tex] = B cos(10°) = 18.3m

In conclusion, The x-component for this vector is : [tex]B_{x}[/tex] = 18.3m

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Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

Answer:

C. Table

A organized collection of records is called a database table. The main key that uniquely identifies each row in the table. Since no two rows may have the same instantaneous value thanks to a unique key, it is simple to choose a row by its main key. A table can also contain several columns, each of which can hold a distinct kind of data. The information stored is verified using the column name and data type.  

The table name is the linguistic component of a database table. It can represent expressions or variable items, and it must be typed exactly. Square brackets denote a need for the item, whereas curly braces denote an optional element. An alternate choice is indicated by a vertical bar. A main key is a single field, and a secondary key might be a combination of many fields. This kind of information is crucial.

Any field in the data table can be referenced by the secondary key. A database's data field may be a string, a number, a date, or a time. A table's first field should always be the main key, followed by any foreign keys. Actual data should never be present in a primary key. Since the immediate key value is derived from external sources, it is crucial to use it properly. Otherwise, data input mistakes may occur.

The information kept in the database table is referred to as a secondary key. This kind is a primary key, which implies it is a secondary key. A primary key can be used as an alternate key to distinguish between different records in a table and to uniquely identify a record inside a database. Multiple fields that can be either main or alternative make up a composite key.

Depending on the type of data being stored and its intended use, a database table may have a single row or several. The main key determines the number of rows, which might be 50 or more. A secondary key can be used to establish connections between various tables and to distinguish between various data types in a database. A data table also includes a secondary key, which is essential for data retrieval and is separate from the primary key.

You can have local or global database tables. The two types of tables differ in terms of names, availability, and visibility. A global table is accessible to all users and may be utilized by any other user, while a local table is only visible to the present user. It is distinct from a typical table and requires more effort to build than a straightforward one does. Its main function is to maintain a database and store data.

Rows in a database table can be of any size. A table's rows can include any number of columns. A database table's rows can be arranged in any way. The number of rows on a data page might range from one to thousands. Each row may consist of a clustered index, a row, or both. A list of columns may also be present in a row. A clustered index can only contain information in a particular sequence.

A table can be set up in  two different ways. The information-containing row of data is its main key. The data column that contains the values for a row is known as a secondary key. A database that employs rows is the third form of table. Multiple primary keys are possible for a data row, and a secondary key is a secondary key that refers to a row column. A database will have a foreign key if it is formed with a single primary key.

There is no limit to the amount of rows and columns in a database table. Up to 1,024 rows are allowed in a typical user-defined table. The server's storage capacity determines how many rows can fit in a database table. A table can have characteristics added to it in addition to the column and row. A restriction can stop a row from having empty fields. The connection between two tables can be established via an important control.

The buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.

The buoyancy needed for an object can be calculated using the formula;

B = ρ × V × g

where;

B = 6000 × 0.0009 × 9.81

B = 52.97N

Therefore, the buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.

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