The total energy need during pregnancy is normally distributed, with mean Mu = 2600 kcal/day and standard deviation = 50 kcal/day. What is the probability that a randomly selected pregnant woman has an energy need of more than 2625 kcal/day? ...?

Answer:

D. surface wave

Explanation:

A) Transverse Waves

As the energy is propagated through the medium and then medium particles moves to and fro perpendicular to wave propagation then this is known as transverse waves.

B) Longitudinal waves

As energy is propagated through the medium and then medium particles oscillates in direction of wave propagation then it is known as longitudinal waves.

C) Primary waves

these waves are due to compression and expansion of medium and it can travel at high speed. It is moreover similar to longitudinal waves

D) Surface Waves

In these type of waves the disturbance or energy is propagated through the medium particles on the surface of the water in the shape of concentric circles.

So surface wave is characterized by circular motion

Answer:

a = 0

Explanation:

When a baseball is thrown straight upward, it moves upward under the action of gravity. We know that at the highest point, the speed of the object is equal to 0. The rate of change of velocity is called acceleration of an object.

Since, the speed is 0 at the height point, so, its acceleration at its greatest height is 0. Hence, this is the required solution.

When a baseball is thrown straight upward, its acceleration at its greatest height is equal to: c. 0 [tex]m/s^2[/tex]

In Physics, the acceleration of an object or body is calculated by subtracting its initial velocity from the final velocity and dividing by the time.

Mathematically, acceleration is given by the formula;

[tex]Acceleration = \frac{V\; - \;U}{t}[/tex]

Where:  

A baseball that is thrown thrown straight upward, experiences motion due to the acceleration of gravity.

Also, the final speed of an object is equal to zero (0) at the greatest height because it came to a stop while the initial velocity is equal to zero (0), since it would accelerate from rest.

Deductively, the rate of change of velocity with respect to time is equal to zero (0) and this is the value of baseball's acceleration.

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The final speed of the 100-g ball is about 5.3 m/s to the left

The final speed of the 200-g ball is about 1.7 m/s to the right

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of ball 1 = m₁ = 100 g

initial velocity of ball 1 = u₁ = 4.0 m/s

mass of ball 2 = m₂ = 200 g

initial velocity of ball 2 = u₂ = -3.0 m/s

Asked:

final velocity of ball 1 = v₁ = ?

final velocity of ball 2 = v₂ = ?

Solution:

Firstly , we will use Conservation of Momentum Law as follows:

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

[tex]100(4.0) + 200(-3.0) = 100v_1 + 200v_2[/tex]

[tex]-200 = 100v_1 + 200v_2[/tex]

[tex]-2 = v_1 + 2v_2[/tex] → Equation 1

[tex]\texttt{ }[/tex]

If the collision is perfectly elastic , then:

[tex]u_1 - u_2 = v_2 - v_1[/tex]

[tex]4.0 - (-3.0) = v_2 - v_1[/tex]

[tex]7.0 = v_2 - v_1[/tex] → Equation 2

[tex]\texttt{ }[/tex]

Let's solve the two Equations above:

(Equation 1) + (Equation 2) ↓

[tex]-2 + 7.0 = (v_1 + 2v_2) + (v_2 - v_1)[/tex]

[tex]5.0 = 3v_2[/tex]

[tex]v_2 = 5.0 \div 3[/tex]

[tex]v_2 = 1\frac{2}{3} \texttt{ m/s}[/tex]

[tex]v_2 \approx 1.7 \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]-2 = v_1 + 2v_2[/tex]

[tex]-2 = v_1 + 2(1\frac{2}{3})[/tex]

[tex]v_1 = -2 - 3\frac{1}{3}[/tex]

[tex]v_1 = -5\frac{1}{3} \texttt{ m/s}[/tex]

[tex]v_1 \approx -5.3 \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

the independent varible is old and new and the dependent variable is 20 T-shirts can I get brainliest

Explanation:

An independent variable is a variable whose value does not depend on any other factor. Whereas a dependent variable is a variable whose value depends on another factor.

Hence, in the given situation old and new detergent are independent variables as they do not depend on any other factor.

On the other hand, 20 dirty T-shirts are dependent variable because they get clean only if the detergent is strong enough.

The atom underwent alpha decay, emitting 2 protons and 2 neutrons, transforming from polonium (Po-170) to lead (Pb-166).

The atom originally was a polonium (Po) atom with 84 protons and 86 neutrons (Po-170). It underwent alpha decay, a type of radioactive decay where it emitted an alpha particle, consisting of 2 protons and 2 neutrons. As a result, the atom transformed into a lead (Pb) atom with 82 protons and 84 neutrons (Pb-166).

This process reduced the atomic number by 2 and the mass number by 4, leading to the formation of a different element. Alpha decay occurs when an unstable nucleus emits an alpha particle, reducing both its atomic number and mass number. It is a natural process observed in certain heavy and unstable elements as they strive to achieve a more stable configuration, often resulting in the formation of a different element.

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For this case what we must do is calculate the centripetal force of the object.

By definition, the centripetal force is given by:

[tex] F = m(\frac{v^2}{r})
[/tex]

Where,

v: tangential speed

r: radius of the circular path

m: mass of the object

Substituting values we have:

[tex] F = 0.25 * (\frac{12.2 ^ 2}{1.6})

F = 23.26
[/tex]

Answer:

The centripetal force of the ball is given by:

[tex] F = 23.26 N [/tex]

Answer:

The answer is B) 23 N

Explanation:

I got the correct answer on the pre-test.

The solution to arctan[tan(7π/4)] is -(π/4).

To solve the expression arctan[tan(7π/4)], simplify it using the properties of trigonometric functions and inverse trigonometric functions.

First, examine the value of tan(7π/4). The tangent function has a period of π, meaning it repeats its values every π radians.

Therefore, find an equivalent angle within the range of one period (0 to π) that has the same tangent value.

7π/4 can be reduced to 3π/4 by subtracting 2π, as it falls within the same period.

Now, arctan[tan(3π/4)].

The arctan function (or inverse tangent) "undoes" the tangent function, giving us the original angle. However, it's important to note that arctan has a range of -π/2 to π/2.

The angle 3π/4 falls within the range of π/2 to 0, so its corresponding angle within the range of arctan is -(π/4).

Therefore, the solution to arctan[tan(7π/4)] is -(π/4).

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The correct answer is A. Sediment

Answer: A. Sediment

Eutrophication is a phenomena in which a water body gets enriched with minerals and nutrients frequently entered into water body due to run-off from the land, which facilitates dense growth of plants especially on the superficial layers of water body. Decaying organisms, fertilizers and pollution all add nutrients and minerals necessary for plant growth specifically remain floating on the superficial layers of water where plant growth takes place but sediments gets settled at the bottom of the water body, the minerals present in it remain settle at the bottom and hence, does not contribute to eutrophication.

The answer is A. Hope this helps! :)

Yes, the stuntman can jump from one building to the other building.

Further explanation:

The stuntman jump from one building inclined at some angle to the other traversing a parabolic path.

Given:

The velocity of stuntman is 5m/s.

The distance between the buildings is 4m.

The difference in the height of the buildings is 2.5m.

The angle of inclination is [tex]{15^ \circ }[/tex].

Concept used:

When stuntman jumps from the top of one building to the top of other building he start running first then he jump at a velocity of 5m/s inclined at an angle of [tex]{15^ \circ }[/tex] from the horizontal of first building.

The velocity of stuntman has two components [tex]{v_x}[/tex] and [tex]{v_y}[/tex], respectively in the X-direction and in the Y-direction.

The expression for the distance in horizontal direction is given as.

[tex]x = \left( {v\cos \theta } \right)t[/tex]

Rearrange the above expression for time.

[tex]t = \dfrac{x}{{\left( {v\cos \theta } \right)}}[/tex]                             …… (1)

Here, t is the time of flight, v is the velocity of object and [tex]\theta[/tex]  is the angle of inclination.

The expression for the distance in Y-direction is given by the second equation of motion.

[tex]y = ut + \frac{1}{2}\left( { - g} \right){t^2}[/tex]                        

Here, u is the velocity in Y-direction and (–g) is the acceleration due to gravity directed in the downward direction.

The expression for the component of velocity in Y-direction is given as.

[tex]u = v\sin \theta[/tex]

Substitute [tex]v\sin\theta[/tex] for u in the above expression.

[tex]y = \left( {v\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}[/tex]                       …… (2)

Substitute 5m/s for v, 4m for x and [tex]{15^ \circ }[/tex] for   in equation (1).

[tex]\begin{aligned}t&=\frac{{4\,{\text{m}}}}{{\left({\left( {5\,{\text{m/s}}}\right)\left( {\cos {{15}^ \circ }}\right)}\right)}}\\&=0.828\,{\text{s}}\\\end{aligned}[/tex]

Substitute [tex]0.828\,{\text{s}}[/tex] for t, 5m/s for v, [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for g and [tex]{15^ \circ }[/tex] for [tex]\theta[/tex] in equation (2).

[tex]\begin{aligned}y&=\left( {\left( {5\,{\text{m/s}}} \right)\sin \left( {{{15}^ \circ }} \right)} \right)\left( {0.828\,{\text{s}}} \right) + \frac{1}{2}\left( { - 9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right){\left( {0.828\,{\text{s}}} \right)^2}\\&=1.071 - 3.36 \\&=- 2.29\,{\text{m}}\\\end{aligned}[/tex]

Thus, the stuntman can jump from one building to another because its vertical distance is less than the difference in height of buildings.

Learn more:

1.  Motion under force https://brainly.com/question/6125929.

2.  Projection of ball https://brainly.com/question/11023695.

3. Conservation of momentum https://brainly.com/question/9484203.

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Force, motion, parabolic path, acceleration due to gravity, time of flight, angle of inclination, inclined plane, buildings, height difference, cliff, horizontal direction, vertical direction, 2.29 m, 2.3m, 0.828sec.

The question involves calculating whether a stuntman can jump from one rooftop to another using principles of projectile motion. By calculating the horizontal and vertical components of the stuntman's velocity, and the time and distance he will travel, we determine he will fall approximately 2.9 meters. Since the other roof is only 2.5 meters lower, the stuntman will successfully make the jump.

The student's question involves determining whether a stuntman can safely jump from one building to another, given his initial velocity, the angle of his jump, and the distance and height difference between the buildings. This is a classic physics problem involving projectile motion, where we consider the horizontal and vertical components of the motion separately. We can ignore air resistance and use the kinematic equations to solve this problem.

Firstly, we need to find the horizontal and vertical components of the initial velocity. The horizontal component, vx, is v × cos(θ), and the vertical component, vy, is v × sin(θ), where v is the initial velocity and θ is the angle of launch. For a jump of 5.0 m/s at 15°:

The time, t, it takes to travel horizontally across the 4.0 m gap is found using t = d / vx, where d is the horizontal distance:

t = 4.0 m / 4.83 m/s ≈ 0.83 s

Now, let's determine if he will drop less than 2.5 m in this time frame. We use the equation of motion for vertical displacement, Δy = vyt + ½gt², with g being the acceleration due to gravity (9.81 m/s²) to find the vertical drop:

Δy = (1.29 m/s × 0.83 s) + (0.5 × -9.81 m/s² × (0.83 s)²) ≈ -2.9 m

The stuntman will fall approximately 2.9 m, thus he will make it to the other roof, which is 2.5 m lower than his initial jump point.

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Answer:

t = 0.55 s and t = 1.14 s

Explanation:

as we know that height in terms of time is given as

[tex]h = 3 + 27t - 16 t^2[/tex]

now we have to find the time at which its height is h = 13 ft

so here we can say

[tex]13 = 3 + 27 t - 16 t^2[/tex]

[tex]16 t^2 - 27 t + 10 = 0[/tex]

[tex]t = \frac{27 \pm \sqrt{27^2 - 4(16)(10)}}{2(16)}[/tex]

[tex]t = \frac{27 \pm 9.43}{32}[/tex]

[tex]t = 1.14, 0.55[/tex]

so it will be at h = 13 ft height at t = 0.55 s and t = 1.14 s

Answer:

E₃ = energy in third energy level.

Explanation:

The energy of an electron is third orbit is given by :

[tex]E=-\dfrac{13.6Z^2}{n^2}[/tex]

Where

Z is the atomic number

n is principal quantum number and n = 1,2,3....

i.e. the energy of an electron is inversely proportional to the principal quantum number.

E₁,E₂,E₃.......... shows the energy of an electron in first, second and third energy level respectively.

Answer: Option A.

Explanation: When you turn on your laptop, you are allowing the electrical energy to flow in the components of the laptop.

Once it is on, you can see that your laptop is generating light (from the screen) and sound (maybe if you are listening to music or something). And as the electricity is passing through the components of the laptop, the components start to raise their temperature, and then you also are generating thermal energy.

The units can be best determined by expressing them both in terms of SI base units.

The Système International (SI) unit is named after the French word for it. The International System of Units (ISO) is the name of the metric system that is used as the industrial standard for measurements (SI). SI units are crucial for the growth of science and technology.

It is composed of 7 base units, from which 22 derivative units are generated. Either a standard multiple or a fractional quantity can be used to express SI units. Prefix multipliers with powers of 10 in the range from [tex]10^{-24}[/tex]  To [tex]10^{24}[/tex]  Are used to define these numbers.

Now, according to the question :

J=kg×m²/s²

J/m²=kg/s²

and, N=kg×m/s²

⇒m×s²/(kg×m/s²)

=[tex]s^{4}[/tex]/kg

Hence, after comparing both of them, we can say that they are not equivalent.

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Answer:

A. Mark's physics book is on the desk. Mark pushes horizontally (from the side) on the book, but the book does not move.

DIAGRAM B

B. Kendall's physics book is on the desk. Kendall pushes horizontally (from the side) on the book, and the book moves across the desk at a constant speed. |

DIAGRAM B

C. Jenny pushes a cart at constant velocity down the grocery store aisle.

DIAGRAM B

D. Eddy is driving his Camaro down a straight, level road while on cruise control.

DIAGRAM B

E. Ryan is accelerating his Dakota down a straight, level road.

DIAGRAM C

F. Jackie's Viper is parked on the shoulder of the road while she changes a flat tire.

DIAGRAM A

Explanation:

A. Since the book does not move even after being pushed it is leveled in all forces.

B. Since it is pushed at a constant speed, all forces are equal.

C. Since it is pushed at a constant speed, all forces are equal.

D. Since it is driven at a constant speed on cruise control, all forces are equal.

E. Since he is accelerating, which means it is being driven much faster than the friction that is pushing the car back, the forces are unequal.

F. The only forces acting on the is the car on the ground and the ground on the car, there is no other forces of friction pushing it back or the car being pushed forward. It is parked.

In option A, the book does not move, so the force diagram would show the normal force and the force of friction. In option B, the book moves at a constant speed, so the force diagram would show the normal force and the force of friction in the same direction as the applied force. Option C involves a cart moving at constant velocity, which would have a force diagram showing the forward push and the force of friction. Options D and E involve cars on level roads, with D having a balanced force diagram and E having unbalanced forces causing acceleration. Option F involves a parked car with a balanced force diagram.

The force diagram that best represents the scenario described in option A, where Mark's physics book is on the desk and Mark pushes horizontally on the book but it does not move, would be a free-body diagram showing only the normal force and the force of friction acting on the book. Since the book does not move, the forces are balanced, with the force of friction equal in magnitude and opposite in direction to the horizontal force applied by Mark.

In option B, where Kendall's physics book is on the desk and Kendall pushes horizontally on the book and it moves across the desk at a constant speed, the force diagram would also include the normal force and the force of friction, but the force of friction would be in the same direction as Kendall's applied force, since the book is moving. The forces are balanced and the net force is zero.

In option C, where Jenny pushes a cart at constant velocity down the grocery store aisle, the force diagram would include a force in the forward direction representing Jenny's push, as well as the force of friction opposing the motion. The forces are balanced and the net force is zero.

In option D, where Eddy is driving his Camaro down a straight, level road while on cruise control, there would be a force diagram showing the force of the engine pushing the car forward, the force of friction opposing the motion, and the normal force. The forces are balanced and the net force is zero.

In option E, where Ryan is accelerating his Dakota down a straight, level road, the force diagram would include the force of the engine pushing the car forward, the force of friction opposing the motion, and the normal force. The net force is non-zero, as the forces are unbalanced and causing the acceleration.

In option F, where Jackie's Viper is parked on the shoulder of the road while she changes a flat tire, there would be a force diagram showing only the force of gravity acting on the car, the normal force from the ground, and the force of friction from the ground. The forces are balanced and the net force is zero.

Accepting the null hypothesis when the experimental hypothesis is true is called type error.

The type error object denotes a failure to perform an operation.When a value is not of the anticipated type.

When an operand or argument provided to a function is incompatible with the type required by that operator or function,the condition of type error is found.

Accepting the null hypothesis when the experimental hypothesis is true is called type error.

Hence,type error is the correct answer for the blank.

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Answer : [tex]L=66.18\ kgm^2/s[/tex]

Explanation :

It is given that,

Mass, m = 5.7 kg

radius, R = 0.43 m

frequency, f = 20 rev/s

Angular momentum, [tex]L=I\omega[/tex]

I is the moment of inertia of disk, [tex]I=\dfrac{MR^2}{2}[/tex]

we know that, [tex]\omega = 2\pi \nu[/tex]

so, [tex]L=MR^2\pi \nu[/tex]

[tex]L= 5.7\ kg\times (0.43\ m)^2\times3.14\times 20\ rev/s[/tex]

[tex]L=66.18\ kgm^2/s[/tex]

Hence, the magnitude of angular momentum of spinning disk is [tex]66.18\ kgm^2/s[/tex].

Answer: 3150 J

Explanation:

Work is the result when a force acts on an object and moves it by some distance and it is calculated in Joules.  

So,

Distance covered by the box = 35m

Force applied on the box = 90N

Formula:

The force and the distance covered was in the same direction, so the angle between them is 0°. The work can be found using the formula:

W = Fd cosθ

W = Fd cos0  

W = Fd(1)

W= (90) * (35)

W = 3150 J

was the answer even right

Answer;

-Reference point

The x represents the reference point on a motion map.

Explanation;

-Motion maps are another way to represent the motion of an object.  (other representations are graphical and mathematical models).

-Motion maps are a visual way to represent an object’s motion at various time

-. A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point.

-The relationships between motion and reference point is that When an object changes position over time relative to a reference point, the object is in motion. This means that there relationship is they are similar.

The x represents a reference point on a motion map. The correct option is A.

Another technique to depict an object's motion is motion maps. Models in mathematics and graphics are alternative forms of representation. A visual representation of an object's motion through time is called a motion map.

A reference point is a location or object that is used as a point of comparison to ascertain whether something is moving. When an object shifts in relation to a fixed point, it is said to be in motion.

The relationship between motion and reference point is that an object is in motion when its position with respect to a reference point varies over time. This indicates that they have a similar relationship.

Therefore, the value x represents a reference point on a motion map. The correct option is A.

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the answer to the question is A.

Answer:

a

Explanation:

C. Acceleration Increases As Mass Decreases.