Which of the following is an example of an offensive action?

Answer:

D

Explanation:

momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.

and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.

so, the sign is only for the direction.

the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.

It is possible that momentum can be negative if an object is moving backward. The correct option is D.

A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.

When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.

Thus the correct option is D.

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The average kinetic energy of the gas's molecules directly relates to its temperature. Faster moving particles will more frequently and violently hit the container walls. Because of this, the force acting on the container's walls increases, increasing the pressure.

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When the volume of gas increased the KE of gases particles increases and hit the walls of container which increases the pressure on the walls .

What if gas will increase?

The gas expands in a closed container . The molecules strike and hit each other therefore the pressure at the walls increase.

According to ideal gas law , when volume is cnstant the Pressure and temperature both will increase.

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Answer:

VR = Velocity of effort / Velocity of Load

Explanation:

The ratio of the out-lever to the in-lever (length of the resistance arm to effort arm) is known as the velocity ratio (VR). It indicates that the distance covered by effort is four times that covered by the load.

The load-to-effort ratio of a machine, or alternatively, the output-to-input ratio of a machine, is known as its mechanical advantage. Another definition of velocity ratio is the ratio of the velocity of effort to the velocity of load.

When a machine has a force ratio of 4 and a velocity ratio of 4, this indicates that the weight moved is multiplied by 4 and the distance moved by the effort is multiplied by 4 at the same time.

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Answer:

Explanation:

the value is -8 cm

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

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Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

[tex]\Sigma \tau = 0[/tex]

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]

Doing the summation using their respective lever arms:

[tex]0 = L Tsin\theta - dF_g[/tex]

[tex]dF_g = LTsin\theta[/tex]

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]

Now, let's solve for 'T'.

[tex]T = \frac{dMg}{Lsin\theta}[/tex]

Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]

Answer:

pictures please

Explanation:

I need a picture so I can tell you

The volume of the helium balloon in order to lift the weight is 17,760m³.

To find the answer, we need to know about the buoyant force.

     => V= 3179/0.179 = 17,760m³

Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.

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The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.

we know:-

Mass = 69 kg

Speed = 3.2 m/s

Coefficient of friction ( ratio of friction force and normal force ) = 0.70

Acceleration due to gravity, g = 9.8 m/s^2

(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:

[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]

      [tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]

      [tex]= 353.28[/tex] Joules

Mechanical energy = 353.28 Joules

(b) To determine how far (distance) the runner slide:

acceleration = ug

                     [tex]= 3.2[/tex] × [tex]9.8[/tex]

                     [tex]= 31.36[/tex] [tex]m/s^{2}[/tex]

distance ,

[tex]V^{2} = U^{2} + 2aS[/tex]

[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]

[tex]10.24 = 62.72 S[/tex]

[tex]S = {\frac{10.24}{62.72} }[/tex]

Distance, S = 0.16 m  

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The horizontal distance XQ traveled by the bomb is 250 m.

Let the XQ be the horizontal distance traveled by the bomb.

h = vt + ¹/₂gt

Let the vertical velocity = 0

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 100 / 9.8)

t = 4.5 s

XQ = vx(t)

where;

vx is horizontal speed, = 200 km/hr = 55.56 m/s

XQ = 55.56 x 4.5

XQ = 250 m

Thus, the horizontal distance XQ traveled by the bomb is 250 m.

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A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) Collecting dew from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

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The percent difference between two numbers [tex]x[/tex] and [tex]y[/tex] is given by

[tex]\dfrac{|y-x|}x \times 100\%[/tex]

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from [tex]x[/tex] to [tex]y[/tex] or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

[tex]\dfrac{|6-3|}3 \times 100\% = 100\%[/tex]

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

[tex]\dfrac{3-6}3\times100\%=-50\%[/tex]

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take [tex]x[/tex] from the left column and [tex]y[/tex] from the right column.

[tex]\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%[/tex]

[tex]\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%[/tex]

[tex]\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%[/tex]

The percentage difference are 2.82%, 4.17%, 4.17%.

The percent difference between two numbers a and b is given by formula:

[tex]\frac{y-x}{x}*100[/tex]

We are given absolute value as we are only concerned about the absolute percent difference.

Making comparisons to object measurements determines that the differences should be computed relative to object measurements.

Here take from the left column and  from the right column.

a) [tex]\frac{7.3-7.1}{7.1} *100=2.82[/tex]%

b) [tex]\frac{5.0-4.8}{4.8} *100=4.17[/tex]%

c) [tex]\frac{7.5-7.2}{7.2} *100=4.17[/tex]%

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Answer:

See below

Explanation:

To change C to F°

  F = 9/5 C   + 32

     = 9/5 ( 45) + 32 = 113° F

To change F to C°   <= you could use the same equation...or re-earrange to:

C = 5/9 (F-32)

   = 5/9 ( 98.6 - 32) = 37° C

Similarly

100 C = 212 ° F

 to change C °  to  K    add 273.15   =  373.15 K    

"V = IR" is the equation that relates current to voltage.

Relationship between current and voltage:

                                           V = IR or,

                                           I = V/R,

        the amount of current (I) flowing through a circuit is inversely    

        related to the amount of resistance (r) and directly proportional to  

        the voltage (v).

Therefore the correct answer is option C i.e., "V = IR".

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Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N

The two conditions are;

The given parameters are;

Tsinθ = mg

Tsin 66 = 21 x 9.8

T = 205.8 / 0.914

T = 225.3 N

Therefore, the tension in the cable is 225.3 N

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The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                                    [tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]

                                 [tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]

                                 [tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.

We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.

                              [tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]

                                [tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]

                                     [tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]

Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is -16.04 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

F(net) = F - μFf

F(net) = 299 - 0.351(92 x 9.8)

F(net) = -24.67 N

W = F(net) x L

W = -24.67 x 0.65

W = - 16.04 J

a = F(net)/m

a = -24.67/92

a = - 0.268 m/s²

v² = u² + 2as

v² = 0.88² + 2(-0.268)(0.65)

v² = 0.426

v = √0.426

v = 0.65 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is -16.04 J.

The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

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Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

From the calculation, the normal force is 6161.2 N.

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

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The kinetic energy of the projectile at the highest point of its motion will be E/4.

When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.

Given, the angle of projection  with horizontal, [tex]\theta = 60 ^o[/tex]

Consider that 'E' is the initial value of the kinetic energy of the projectile.

The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]

where m is the mass of the given projectile.

The component of the velocity of the projectile in the horizontal direction:

uₓ = u cosθ

uₓ = u cos 60°

uₓ = u/2

From the equation of motion: v = u +at

v = (u/2) + (0) t

v = u/2

The final kinetic energy of the projectile:

[tex]E_f = \frac{1}{2}mv^2[/tex]

[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]

[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]

[tex]E_f = \frac{E}{4}[/tex]

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Answer:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

Accuracy means the state of being accurate, without any mistakes and the results should be 100% true.

Whereas, precision means, approximately true or almost true.

Hope it help you

The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (40 x 9.8)/(2 x sin31)

T = 380.55 N

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

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The stiffness constant of the spring is 68,290.3 N/m

Apply the principle of conservation of energy;

U = K.E

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

where;

k = (1300 x 16.67²)/(2.3²)

k = 68,290.3 N/m

Thus, the stiffness constant of the spring is 68,290.3 N/m.

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The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as

B=1.9*10^{-5}T

To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

Generally, the equation for Biot savant law is  mathematically given as

[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Their net field is

Bn=2B

[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Hence

[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]

B=1.9*10^{-5}T

In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

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Answer:

The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.

Explanation:

The general form of a similar question to this is:

[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]

where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)

This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.

The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].

Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.

Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].

This gives us our new equation for the Magnetic Field produced by a single segment at a point:

[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]

Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!

Now we can start plugging things in:

[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]

[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]

This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.

Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:

[tex]B = 1.57*10^{-5} T[/tex]; into the page.

Hello!

a)

To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.

Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)

i = Current in wire ('I' A)
dl = differential length element

r = distance from wire to point P ('R' m, remains constant!)

We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

First, let's mind the cross-product.

The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.

Therefore, we can disregard the cross-product since sin(90) = 1.

Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:

[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]

We have a semi-circle, so we are integrating from 0 to π radians.

[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]

Simplifying to make the integral easier, we can take constants out of the integral.

[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]

Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]

b)

Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.

Answer:

Higher levels of organization are built from lower levels.

Molecules combine to form cells.

Cells combine to form tissues.

Tissues combine to form organs.

Organs combine to form organ systems,

and organ systems combine to form organisms.

Explanation:

Hope it helps.

The elastic potential energy of the block-spring system is 0.906 J.

Velocity of the block , v = 0.83 m/s.

Elastic potential energy is the energy stored in a stretched or compressed elastic material.

The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J

The elastic potential energy of the spring is converted to kinetic energy of the block.

1/2 mv² = 0.906 J

where v is velocity

v = √(0.906 * 2)/2.6

v = 0.83 m/s.

In conclusion, elastic potential energy is present in compressed or stretched elastic materials.

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Big bang happened about 13.7 billion years ago in our universe.

According to the big bang theory, about 13.7 billion years ago, an explosive expansion began, expanding our universe outwards faster than the speed of light.

According to the flat model, the universe is infinite and will continue to expand forever because the universe is expanding.

Cosmic background radiation is a weak radio-frequency radiation that is traveling through outer space in every direction. It is the residual radiation of the big bang, when the universe was very hot.

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe has all over radiation which is called the “cosmic microwave background".

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The the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

The lead weight will fall under the influence of gravity with magnitude of 9.8 m/s².

t = √2h/g

where;

t = √((2 x 6.004)/9.8)

t = 1.11 s

Thus, the the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

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The spring constant, k, in newtons per meter is 1,955.9 N/m.

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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