A cube, whose mass is 0.660 kg, is attached to a spring with a force constant of 106 N/m. The cube rests upon a frictionless, horizontal surface (shown in the figure below).

The cube is pulled to the right a distance A = 0.140 m from its equilibrium position (the vertical dashed line) and held motionless. The cube is then released from rest.

(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the cube?
________N



(b) At that very instant, what is the magnitude of the cube's acceleration (in m/s2)?

_____ m/s2

(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
-Away from the equilibrium position (i.e., to the right in the figure).
- Toward the equilibrium position (i.e., to the left in the figure).
- You cannot tell without more information.


***Please refer to attached photo

The upward force exerted on the board by the support is 530.8 N.

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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Answer:

oa

Explanation:

it may be oa is the right answer for this question

but I don't know properly

Answer:

A. the center of mass

Explanation:

[tex]C_m[/tex] is usually used to designate the center of mass of an object or system of objects.

B. is incorrect because the combined mass of a system is usually denoted by ∑m.

C. is incorrect because the momentum of a system is denoted by the letter p.

(1) The normal force exerted by the road on the car at point A is 1.06 x 10⁴ N

(2) The normal force exerted by the road on the car at point B is 7,495.8 N.

Fn(A) = mg

where;

Fn(A) = 1080 x 9.81

Fn(A) = 1.06 x 10⁴ N

The normal force exerted by the road on the car at point B is calculated as follows;

Fn = Fn(A) - mv²/r

where;

Fn = 1.06 x 10⁴ - (1080 x 19.33²)/(130)

Fn = 7,495.8 N

Thus, the normal force exerted by the road on the car at point A is 1.06 x 10⁴ N and the normal force exerted by the road on the car at point B is 7,495.8 N.

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a)The velocity of the knife relative to the ship when it hits the deck of the ship will be 14.00 m/sec.

b)The velocity of the knife relative to a stationary observer on shore is 14.10 m/sec

c)The position at which the knife hits the deck is 82.8°.

The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

The velocity of the knife relative to the ship when it hits the deck of the ship is;

v=√2gh

v=√2×9.81 ×10.0 m

v=14.00 m/sec

b)

The velocity of the knife relative to a stationary observer on shore is found as;

v₁₂ = √[(14.00)²+(1.74)²]

v₁₂=14.10 m/sec

c)

The position at which the knife hits the deck is;

[tex]\rm \theta = tan^{-1}\frac{14.00}{1.75} \\\\ \theta = 82.8 ^ 0[/tex]

Hence, the velocity of the knife relative to the ship, the velocity of the knife relative to a stationary observer on shore, and the position at which the knife hits the deck will be 14.00 m/sec,14.10 m/sec, and 82.8° respectively.

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The upward force exerted on the board by the support is 530.8 N.

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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The prompt above requires an explanatory essay. The focus of an explanatory essay is to give clarity to a concept. See the sample below.

From history, it is clear that Science builds on itself. For avoidance of doubt, this means that improvements in science are incremental with each improvement feeding off the last.

From the lesson we have examples of this phenomena such as that of Galileo. Recall that the text indicates that He improved the telescope; and while carrying out the same research that his predecessor Copernicus had done in relation to the stars and planets, Galileo used math to justify his position.

It is also on record that Johannes Kepler who lived in the same period as Galileo took interest in Astronomy. In this case, science built on itself because, Kepler's work served as the foundation for Sir Isaac Newton's work. This is also another example of how science builds on itself.

From the above textual evidence, it is given beyond reasonable double that science indeed builds on itself with the discovery of each generation serving as the foundation for the advancement of the next.

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The workdone required to stack them over each other is 39.69 J

This is defined as the product of force and distance moved in the direction of the force.

Workdone (Wd) = Force × distance (d)

But

Force = weight = mass × acceleration due to gravity

w = mg

Distance (d) = height (h)

Thus,

Wd = weight × height

We'll begin by calculating the weight of each bricks

W = mg

W = 1.5 × 9.8

W = 14.7 N

Next, we shall determine the work done in stacking the 2nd on the 1st brick and so on..

Workdone in stacking the 2nd on the 1st

Wd = Weight × height

Wd = 14.7 × 0.06

Wd = 0.882 J

Workdone in stacking the 3rd

Wd = Weight × height

Wd = 14.7 × 0.12

Wd = 1.764 J

Workdone in stacking the 4th

Wd = Weight × height

Wd = 14.7 × 0.18

Wd = 2.646 J

Workdone in stacking the 5th

Wd = Weight × height

Wd = 14.7 × 0.24

Wd = 3.528 J

Workdone in stacking the 6th

Wd = Weight × height

Wd = 14.7 × 0.3

Wd = 4.41 J

Workdone in stacking the 7th

Wd = Weight × height

Wd = 14.7 × 0.36

Wd = 5.292 J

Workdone in stacking the 8th

Wd = Weight × height

Wd = 14.7 × 0.42

Wd = 6.174 J

Workdone in stacking the 9th

Wd = Weight × height

Wd = 14.7 × 0.48

Wd = 7.056 J

Workdone in stacking the 10th

Wd = Weight × height

Wd = 14.7 × 0.54

Wd = 7.938 J

Total workdone = 0.882 + 1.764 + 2.646 + 3.528 + 4.41 + 5.292 + 6.174 + 7.056 + 7.938

Total workdone = 39.69 J

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The WEIGHT of each brick is (m g) = 14.7 Newtons

Work to lift each brick = (force x distance) = [14.7 x (distance lifted)] Joules

Place 2nd brick on top of the 1st:  lift 6cm,  (14.7N x .06m) = 0.882 J

Place 3rd brick on top of the 2nd:  lift 12cm,  (14.7N x .12m) = 1.764 J

Place 4th brick on top of the 3rd:  lift 18cm,  (14.7N x .18m) = 2.646 J

Place 5th brick on top of the 4th:  lift 24cm,  (14.7N x .24m) = 3.528 J

Place 6th brick on top of the 5th:  lift 30cm,  (14.7N x .3m) = 4.41 J

Place 7th brick on top of the 6th:  lift 36cm,  (14.7N x .36m) = 5.292 J

Place 8th brick on top of the 7th:  lift 42cm,  (14.7N x .42m) = 6.174 J

Place 9th brick on top of the 8th:  lift 48cm,  (14.7N x .48m) = 7.056 J

Place 10th brick on top of the 9th:  lift 54cm,  (14.7N x .54m) = 7.938 J

Total work to stackum (addum up)  = 39.7 Joules

____________________________________________

Way #2  (easier):

Weight of each brick = (mg) = 14.7 Newtons

Work to stack each one = (14.7N) x (distance lifted) Joules

Number of bricks to lift  =  9

AVERAGE distance to lift them = (1/2) (54cm + 6cm) = 30cm

Average work = (weight of each brick) x (average distance lifted)

Total work = (number of bricks) x (weight of each) x (average lift)

Total work = (9 bricks) x (14.7N) x (0.3m) = 39.69 Joules

The concept of a black hole seem reasonable to me because of its valid and authentic evidences.

A black hole is a region in space where light is unable to escape due to strong gravitation force. The strong gravity is formed because matter has been pressed into a tiny space. This compression occurs at the end of a star's life. Some black holes are formed when a stars die.

So we can conclude that the concept of a black hole seem reasonable to me because of its valid and authentic evidences.

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The time taken for the bus to travel 95.1 m is 2.8 s

v² = u² + 2as

v² = 26.8² + (2 × 4.73 × 95.1)

v² = 1617.886

Take the square root of both sides

v² = √1617.886

v = 40.2 m/s

v = u + at

40.2 = 26.8 + (4.73 × t)

Collect like terms

4.73 × t = 40.2 - 26.8

4.73 × t = 13.4

Divide both side by 4.73

t = 13.4 / 4.73

t = 2.8 s

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The density of the mixture is determined as 1,013.75 kg/m³.

The density of the mixture is calculated as follows;

density = total mass of the mixture / total volume of the mixture

mass of the fresh water = density x volume

                                        = 1000 kg/m³ x 0.0018 m³

                                        = 1.8 kg

mass of the sea water = 1025 kg/m³ x 0.0022 m³

                                     = 2.255 kg

Total mass = 1.8 kg + 2.255 kg = 4.055 kg

Total volume = 0.0018 m³  + 0.0022 m³ = 0.004 m³

Density = 4.055 kg/0.004 m³ = 1,013.75 kg/m³

Thus, the density of the mixture is determined as 1,013.75 kg/m³.

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Answer:

Fruit Punch

Explanation:

A pure substance is whereby there is only one type of element or a compound in the periodic table in the substance.

Carbon: Well its just C in the periodic table, so this is definitely pure.

Water Molecule: What makes water? H2O right? Contains Hydrogen and Oxygen, and as we all know H2O is a compound, therefore this is a pure substance.

Fruit Punch: What makes fruit punch, water and fruits. Fruits may contain citric acid(a compound itself), and is mixed with water with already has a compound, so having 2 different compounds will result in a mixture and therefore it will not be pure.

B. Objects A and C possess charges of opposite sign.

The attraction of two objects occurs when the two objects have opposite charges.

The repulsion of two objects occurs when the two objects have same charges.

Thus, we can conclude that, objects A and C possess charges of opposite sign.

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The depth of the well at the given air temperature is 456.3 m.

The depth of the well is calculated from the principle of echo.

v = 2d/t

2d =  vt

d = vt/2

where;

d = (338 x 2.7)/2

d = 456.3 m

Thus, the depth of the well at the given air temperature is 456.3 m.

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B. The meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

A Greek philosopher  known as Democritus first thought of the existence of tiny particles that compose everything around us.

Democritus of Abdera, named the building blocks of matter atomos, meaning literally “indivisible.

Thus, the meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

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The acceleration is  1.55 m/s^2 while is component that is parallel to the ground is 0.775  m/s^2.

The term acceleration is the rate of change of speed. Given that the distance covered is 35.0 m and the final velocity is 10.43 m/s it the follows that;

v^2 = u^2 + 2as

u = 0 because he started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

The component of this acceleration which is parallel to the ground is;

ax = a sin θ

ax =  1.55 m/s^2 cos 60

ax = 0.775  m/s^2

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The maximum height reached by the human cannonball is 4.74 m.

The height reached is calculated as follows;

H = u²sin²θ/2g

where;

H = (15² x [sin(40)]² )/(2 x 9.8)

H = 4.74 m

Thus, the maximum height reached by the human cannonball is 4.74 m.

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The geographical region where I would expect the warmest weather is: C. in the southeast.

A weather map can be defined as a type of chart that is typically used to provide information about the average atmospheric condition of a particular geographical region over a specific period of time.

Based on the weather map shown in the image attached below, we can infer and logically deduce that the geographical region where the warmest weather is expected is in the southeast due to its very high atmospheric pressure.

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The types of fusion that occurs in stars are as follows:

Nuclear fusion is the process by which nucleus of small atoms combine together to produces atoms of larger nucleus.

The types of nuclear fusion that can occur in a star are;

The proton-proton fusion occur in stars which have core temperatures less than 15 million Kelvin having minimum masses of 0.08 solar mass.

The carbon cycle fusion occurs in stars with core temperatures greater than 15 million Kelvin having minimum masses of 4.0 solar mass.

The helium fusion occurs in stars with core temperatures greater than 100 million Kelvin having minimum masses of 0.5 solar mass.

In conclusion, fusion reactions occurring in the stars produce the energy released by stars.

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The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Force,F = ?

Mass,m = 1350-kg

Acceleration,a = 2.00 ✕ 10⁴ m/s²

The magnitude of the force exerted on the ship by the artillery shell is found as;

F=ma

F=1350-kg × 2.00 ✕ 10⁴ m/s²

F= 10,000 N

The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.

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The angular acceleration of the cylinder is mathematically given as

[tex]\alpha = 60.2175 rad/s^2[/tex]

Question parameters

Generally, the equation for Torque generated in the cylinder is  mathematically given as

[tex]\sigma[/tex] = w×r

for circular motion torque is

[tex]\sigma = × \alpha[/tex]

t[tex]=0.5 × m ×r^2 × \alpha\\\\ =0.5 × 2.15 ×0.111 ^2 × \alpha[/tex]

t = 0.0132 ×[tex]\alpha[/tex]

Therefore

0.0132 × [tex]\alpha[/tex] = w × r

[tex]\\\\\alpha =\frac{ 7.161 0.111}{(0.0132)}[/tex]

[tex]\alpha = 60.2175 rad/s^2[/tex]

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The magnitude of the tension in the string marked A and B is mathematically given as

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x.

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

C=

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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Use the fundamental theorem of calculus.

[tex]f(t) = f(0) + \displaystyle \int_0^t f'(u) \, du[/tex]

In order to find velocity and position exactly, you need to know the initial velocity and position.

1) Integrate the acceleration function to get the velocity function. By the FTC,

[tex]v(t) = v(0) + \displaystyle \int_0^t a(u) \, du[/tex]

[tex]v(t) = v(0) + \displaystyle \int_0^t (3u^3+2u^2) \, du[/tex]

[tex]v(t) = \dfrac34 t^4 + \dfrac23 t^3 + v(0)[/tex]

At [tex]t=2\,\rm s[/tex], the velocity is

[tex]v(2) = \dfrac34 2^4 + \dfrac23 2^3 + v(0) = \boxed{\dfrac{52}3 + v(0)}[/tex]

2) Integrate the velocity function to the get the position function. By the FTC again,

[tex]x(t) = x(0) + \displaystyle \int_0^t v(u) \, du[/tex]

[tex]x(t) = x(0) + \displaystyle \int_0^t \left(\frac34 u^4 + \frac23 u^3 + v(0)\right) \, du[/tex]

[tex]x(t) = \displaystyle \frac3{20} t^5 + \frac16 t^4 + v(0) t + x(0)[/tex]

At [tex]t=4\,\rm s[/tex], the object's position is

[tex]x(4) = \dfrac3{20}4^5 + \dfrac16 4^4 + 4v(0) + x(0) = \dfrac{2944}{15} + 4v(0) + x(0)[/tex]

Fortunately, you don't need the initial position to find the displacement, since [tex]x(0)[/tex] cancels out in the end.

[tex]\Delta x = x(4) - x(0) = \boxed{\dfrac{2944}{15} + 4v(0)}[/tex]

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r   ⇒   v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

Answer:it's 132.2m

Explanation:

Hello !

Because first recall parabolic peak height equation and

given that v=47.4m/s , h=10.5m

secondly recall parabolic total distance equation

The distance from the base of the wall that the cannonball land will be 138.7 m.

Consider that a body of mass 'm' is projected with a velocity of projection 'v' and angle of projection 'θ'. Assume that 'h' be the maximum height attained by the projectile.

2gh = v²sin²θ

This expression can be used to calculate the maximum height of the projectile.

Given, the height of the castle wall, h = 10.5 m

The velocity of the cannonball, v = 47.4m/s

The peak height equation can be used to find the angle θ:

v²sin²θ = 2gh

(47.4)² sin²θ = 2(9.81)(10.5)

Sinθ = 0.3028

θ = 17.63°

Therefore, the distance (d) from the base of the wall that the cannonball land:

d =  (47.4)² × 2 sin(17.63) / 9.81

d = 138.7m

Therefore the distance from the base of the wall that the cannonball land is 138.7m.

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1. The speed (in light-years per year) of the debris is 4.67×10⁻² light-years / year

2. The speed (in Km/h) of the debris is 50436000 Km/h

3. The speed of light is 21.4 times the speed of the debris

Speed is the distance travelled per unit. Mathematically, it can be expressed as:

Speed = distance / time

Speed = distance / time

Speed = 0.7 / 15

Speed = 4.67×10⁻² light-years / year

1 light-years / year = 1.08×10⁹ Km/h

Therefore,

4.67×10⁻² light-years / year = 4.67×10⁻² × 1.08×10⁹

4.67×10⁻² light-years / year = 50436000 Km/h

Comparison = Speed of light / speed of debris

Speed of light / speed of debris = 3×10⁸ / 14010000

Speed of light / speed of debris = 21.4

Cross multiply

Speed of light = 21.4 × Speed of debris

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In a fusion nuclear reactor, a drop in temperature would most likely result in a decline in the rate of energy output.

The utilization of nuclear reactions to generate energy is known as a nuclear power. Nuclear fission, nuclear decay, and nuclear fusion processes are all sources of nuclear energy.

Nuclear power facilities currently produce the great bulk of the electricity generated by nuclear fission of uranium and plutonium.

The environment must be hot enough for the deuterium and tritium ions' kinetic energies to be sufficient to break through the Coulomb barrier and fuse together.

Hence a decrease in temperature would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor.

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Answer:

A: a drop in the temperature in the reactor

Explanation:

right on edg

Assuming players stick together immediately after the collision, their speed will be 1.3206 m/sec, and - ve indicates their direction.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of 1st player= 101.0 kg

(u₁) is the initial velocity of 1st player (u₁) = 4.10 m/s

(m₂) is the mass of 2nd player = 117 kg

(u₂) is the initial velocity of 2nd player= -6  m/s

(v) is the velocity after collision =.?

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)V

101.0 kg × 4.10 m/sec + 117 kg × (-6 m/sec) = (101.0 kg + 117 kg) × V

V= - 1.3206 m/sec

Hence, the velocity of players just after impact, if they cling together, will be - 1.3206 m/sec, and - ve shows the direction.

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