how did the parental or caregiver influence impact the child​

Answer:

24° north of east

Explanation:

tan(x)=90/200 ie x=arctan(90/200)=24°

So the plane took off 24° east.

Answer:

24° north of east

Explanation:

Hope this will help

Answer:

v = 17.14 m/s

Explanation:

Given that,

Mass of eagle is 3.75 kg

We need to find the speed of Eagle after it has fallen freely from the rest through a distance of 15m. We can use third equation of motion :

\(v^2-u^2=2as\)

u = 0 (at rest) a= g

\(v^2=2gs\\\\v=\sqrt{2gs} \\\\v=\sqrt{2\times 9.8\times 15} \\\\v=17.14\ m/s\)

so, the speed of the eagle is 17.14 m/s.

Answer:

this much I think 101.9716212978

Answer:

approximately 102kg

Explanation:

1 kg on the surface of the Earth weighs 9.80665 N

1000 N = 1000/9.80665 kg = 102 kg

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a) Object B has one-quarter the mass of Object A but the same amount of kinetic energy. Therefore, Object B must be moving faster than Object A, as it has less mass to carry the same amount of energy.

b) The kinetic energy of an object is given by:

KE = (1/2)\(mv^2\)

where m is the mass of the object and v is its velocity. We are given that Object A has 24 J of kinetic energy, so we can write:

24 J = (1/2)m_av_\(a^2\)

Similarly, Object B also has 24 J of kinetic energy, but with one-quarter the mass of Object A:

24 J = (1/2)(1/4)m_av\(_b^2\)

Simplifying, we get:

v_b/v_a = sqrt(2)

Therefore, the ratio of the speeds of Object A and Object B is sqrt(2) to 1, or approximately 1.41 to 1.

c) From the work-energy theorem, we know that the work done on an object is equal to its change in kinetic energy:

W = ΔKE

If a total work of -12 J is done on each object, we can write:

-12 J = ΔKE

Using the equation for kinetic energy, we can write:

-12 J = (1/2)m_av_\(a^2\) - (1/2)m_av_{ai\(}^2\)

Simplifying, we get:

-24 J = m_av_a^2 - m_av_{ai\(}^2\)

Since we are interested in the ratio of the final and initial speeds, we can divide both sides by (1/2)m_av_{ai}\(^2:\)

-48 = (v_\(a^2\)/v_{ai\(}^2\)) - 1

Solving for the ratio of the final and initial speeds:

v_a/v_{ai} = sqrt(47)

Therefore, the speed of each object changes by a factor of approximately sqrt(47) if a total work of -12 J is done on each object.

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Full Question ;

Object A Has 24 J Of Kinetic Energy. Object B Has One-Quarter The Mass Of Object A. A)If Object B Also Has 24 J Of Kinetic Energy, Is It Moving Faster Or Slower Than Object A? B)Find The Ratio Of The Speeds Of Objects A And B. C)By What Factor Does The Speed Of Each Object Change If Total Work -12 J Is Done On Each?

Object A has 24 J of kinetic energy. Object B has one-quarter the mass of object A.

a)If object B also has 24 J of kinetic energy, is it moving faster or slower than object A?

b)Find the ratio of the speeds of objects A and B.

c)By what factor does the speed of each object change if total work -12 J is done on each?

When a pendulum in the shape of a rod is held at one end and released from rest at an angle, it begins to oscillate about its equilibrium position.

The motion of the pendulum can be described by the simple harmonic motion equation:

θ = θ0*cos(ωt + φ)

where θ is the angle of the pendulum at time t, θ0 is the amplitude of the oscillation (i.e., the maximum angle from the equilibrium position), ω is the angular frequency of the oscillation, and φ is the phase angle.

The angular frequency of a simple pendulum depends on its length and the acceleration due to gravity.

However, the rod pendulum is a special case where the length is not well-defined, and the motion is affected by the mass distribution of the rod.

Therefore, the angular frequency of the rod pendulum is more complicated and depends on its moment of inertia and the torque due to gravity.

The period of oscillation T, which is the time for the pendulum to complete one full oscillation, is related to the angular frequency by:

T = 2π/ω

The period of a rod pendulum is typically longer than that of a simple pendulum with the same length, due to its more complex motion. However, the exact value of the period depends on the specific geometry and mass distribution of the rod.

In summary, when a rod pendulum is released from rest at an angle, it begins to oscillate about its equilibrium position with a complex motion that depends on the mass distribution of the rod.  The period of the oscillation is related to the angular frequency and is longer than that of a simple pendulum.

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Answer: concave lens

Explanation:

Answer:

A or convex lens

Explanation:

Answer:

A) 0 V

B) -117 kV

C) -0.468 J

Explanation:

q1=+2.00μC  q2=−2.00μC q3 = -4.00 μC

A) The electric potential at point a due to q1 and q2 (\(V_a\)) is given as:

\(V_a=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=r_1=d.\ Therefore\\V_a=k(\frac{q_1}{d}+\frac{q_2}{d} )=k(\frac{2}{d}-\frac{2}{d} )=0\)

B) he electric potential at point b due to q1 and q2 (\(V_b\)) is given as:

\(V_b=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=4.5 cm=0.045m,\ r_1=\sqrt{0.045^2+0.045^2}= 0.0636.\ Therefore\\V_b= k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )= 9*10^{9}(\frac{2*10^{-6}}{0.0636}-\frac{2*10^{-6}}{0.045} )=-117\ kV\)

C) The work done on q3 by the electric forces exerted by q1 and q2 (W) is given by:

\(W=q_3(V_a - V_b)=-4*10^{-6}(0-(-117*10^3))=-0.468\ J\)

Answer: Rotational motion

Explanation: The two types of motion of the rolling ball are translational motion and rotational motion. The center of mass of the ball undergoes translational motion whereas the ball undergoes rotational motion in the center of mass frame.

Answer:

B. Make the work you do feel easier

Answer:

Something a machine can do is Decrease the work needed to do something and a machine can also do make the work you do easier.

Explanation:

Answer: arrange systematically in groups; separate according to type, class, etc.

Explanation:

i hooked this up btw

Answer:

Approximately \(2.8\; {\rm m\cdot s^{-1}}\).

Explanation:

If an object of mass \(m\) is travelling at a velocity of \(v\), the momentum \(p\) of that object would be \(p = m\, v\).

Momentum before collision:

Total momentum before collision:

Momentum after collision:

Momentum is conserved immediately before and after collisions. Hence, the total momentum of the two masses after the collision would be equal to the sum of their momentum before the collision, \(54\; {\rm kg \cdot m\cdot s^{-1}}\).

Subtract the momentum of the \(4.5\; {\rm kg}\) mass after the collision from the sum of momentum after the collision: \(54\; {\rm kg \cdot m\cdot s^{-1}} - 36\; {\rm kg \cdot m \cdot s^{-1}} = 18\; {\rm kg \cdot m \cdot s^{-1}}\).

Hence, the momentum of the \(6.5\; {\rm kg}\) mass after the collision would be \(18\; {\rm kg \cdot m \cdot s^{-1}}\). The velocity of this mass would be:

\(\begin{aligned}v &= \frac{p}{m} \\ &= \frac{18\; {\rm kg\cdot m \cdot s^{-1}}}{6.5\; {\rm kg}} \approx 2.8\; {\rm m\cdot s^{-1}}\end{aligned}\).

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

\(volume = \frac{mass}{density} \\\)

From the question

mass = 24 g

density = 1.2 g/mL

We have

\(volume = \frac{24}{1.2} \\ \)

We have the final answer as

Hope this helps you

Answer:

The five conditions of chemical change: color change, formation of a precipitate, formation of a gas, odor change, temperature change.

The bandwidth of the periodic composite signal is drawn as a range between 100 Hz and 2100 Hz , The data rates for sending a digital signal using one harmonic, three harmonics, and five harmonics on a TV channel with a 6 MHz bandwidth would be 6 MHz, 18 MHz, and 30 MHz .

For the first question

Draw the bandwidth of a periodic composite signal, we need to consider the highest frequency component present in the signal.

We have two sine waves one with a frequency of 100 Hz and the other unspecified. Since the bandwidth is given as 2000 Hz, we can assume that the second sine wave has a frequency of 2100 Hz (2000 Hz above the first sine wave frequency).

Draw the bandwidth, we can create a graph with frequency on the x-axis and amplitude on the y-axis.

We plot the amplitude values for the two sine waves at their respective frequencies (100 Hz and 2100 Hz). The bandwidth will be the range between these two frequencies on the x-axis.

For the second question

The data rate for a digital signal transmitted using one harmonic, three harmonics, and five harmonics can be calculated by multiplying the channel bandwidth by the number of harmonics used. Since the bandwidth is given as 6 MHz, the data rates would be as follows:

One harmonic: 6 MHz

Three harmonics: 18 MHz

Five harmonics: 30 MHz

The data rate increases with the number of harmonics used because each harmonic contributes additional information to the signal, allowing for a higher data transmission rate.

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According to the statement the speed at which the ball hits the ground is v = 14.3 + 9.8 x 1.44 = 28.0 m/s.

To find the speed at which the ball hits the ground, we need to use the concept of projectile motion. When an object is thrown or kicked, it follows a curved path, and its velocity can be broken down into horizontal and vertical components. In this case, the initial velocity of the ball is 10.3 m/s in the horizontal direction and 14.3 m/s in the vertical direction.
The vertical velocity of the ball is affected by the force of gravity, which causes it to accelerate downwards at a rate of 9.8 m/s². Using the equation of motion, v² = u² + 2as, we can find the time it takes for the ball to hit the ground.
The initial vertical velocity is u = 14.3 m/s and the final velocity at impact is v = 0 m/s. The acceleration due to gravity is a = 9.8 m/s² and the distance traveled is s. Therefore, we can rearrange the equation to get s = (v² - u²)/2a.
Substituting the values, we get s = (0² - 14.3²)/2(-9.8) = 10.4 m. This means that the ball travels 10.4 meters in the vertical direction before hitting the ground.
To find the speed at which it hits the ground, we can use the formula, v = u + at, where u is the initial velocity, a is the acceleration due to gravity, and t is the time taken to hit the ground.
We have already calculated the time as t = √(2s/a) = √(2 x 10.4/9.8) = 1.44 seconds.
Therefore, the speed at which the ball hits the ground is v = 14.3 + 9.8 x 1.44 = 28.0 m/s.
In conclusion, the ball hits the ground with a speed of 28.0 m/s, which is the result of its initial velocity of 10.3 m/s in the horizontal direction and 14.3 m/s in the vertical direction, and the acceleration due to gravity in the vertical direction.

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Answer:

71 % eff

Explanation:

if 50 out of 70  watt goes to making light

50/70 = 71 %

Answer:

B.  10 m

Explanation:

f = v/λ = (20 m/s)λ / (5 m) = 4 Hz    original frequency

1/2(4 Hz) = 2 Hz   half the original frequency

(20 m/s) / λ = 2 Hz

λ = (20 m/s) / (2 Hz) = 10 m

The second wavelength would increase to 10 m

Answer:

A) but be sure and read the answer.

Explanation:

If the object does not move at all, (that's an important restriction) the net force = 0. That being so, the acceleration must be 0 as well. But there is no law saying that there cannot be a constant motion and that's why the restriction is important.

For a block undergoing simple harmonic oscillation, the maximum amplitude is mathematically given as

a=0.0749m

Generally, the equation for the maximum acceleration is mathematically given as

Ma=A*w^2

Where

A = 0.075 m

Therefore

0.075 * w^2 = 9.8

w = 11.43 rad/s

In conclusion, maximum frequency

mF=w/2pi

mF=11.43/2pi

mF= 1.82 Hz

Where

g=2\pif)^2a

a=(9.8)/4*\pi^2(1.82)^2

a=0.0749m

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The tension on the wire is 54.45 Newtons.

To calculate the tension on an aluminum wire with a density of 2700 kg/m³, a diameter of 4.6 mm, and a transverse wave speed of 38 m/s, we need to use the formula for wave speed in a string:

v = √(T/μ),

where v is the wave speed, T is the tension, and μ is the linear density of the wire.

First, we need to find the linear density (μ) of the wire. To do this, we will use the formula:

μ = (πd²ρ) / 4,

where d is the diameter and ρ is the density of aluminum.

1. Convert diameter from mm to m:
d = 4.6 mm = 0.0046 m

2. Calculate linear density (μ):
μ = (π * (0.0046 m)² * 2700 kg/m³) / 4
μ ≈ 0.0377 kg/m

Now, we can find the tension (T) using the wave speed formula:

3. Rearrange the formula to solve for T:
T = μ * v²

4. Calculate tension (T):
T = 0.0377 kg/m * (38 m/s)²
T ≈ 54.45 N

The tension on the aluminum wire is approximately 54.45 Newtons.

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Answer:

80%

Explanation:

Answer:

sandstone and mudstone

Explanation:

Answer:

D.Linegraph

Explanation:

Linegraph could be easily converted into a pie chart

Answer:

bar graph

Explanation:

Answer: 40J

Explanation:

Gravitational potential energy is defined as the mass of the object, multiplied by the acceleration due to gravity, and the height from your arbitrary reference point. If we assume that the floor has a gravitational potential of zero, then the gravitational potential of the book on the shelf would be:

\(U=mgh\)

We were given the weight of the book, not the mass. In order to find the mass of the book, we need to divide the weight by the acceleration of gravity. We have that:

\(Mass=Weight/Gravity\)

\(Mass=16N/9.8m/s^2\)

\(Mass=1.63kg\)

Plug everything into the gravitational potential formula to get:

\(U=(1.63kg)*(9.8m/s^2)*(2.5m)\)

\(U=40J\)

The gravitational potential energy of the book placed 2.5 m above from the ground is 40 joules.

The energy of the body due to its position with reference to some other body is called the potential energy of the two - body system.

Given is an object whose weight is 16 N placed on a shelf that is 2.5 m from the ground.

The weight of the book is -

W = mg = 16 N

The gravitational potential energy of the book will be -

E = mgh = 16 x 2.5 = 40 joules

Therefore, the gravitational potential energy of the book placed 2.5 m above from the ground is 40 joules.

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a) The motion of the object between 15 s to 30 s is increasing velocity, to a constant velocity and finally a decreasing velocity.

(b) The average velocity of the object between 0 and 15 seconds is 0.167 m/s.

(c) The position of the object at 5.0 seconds is 0.5 m.

(d) Between 30 and 40 seconds, the velocity of the object is decreasing and the object is decelerating.

(a) The motion of the object between 15 s to 30 s can be described as increasing velocity, to a constant velocity and finally a decreasing velocity.

(b) The average velocity of the object between 0 and 15 seconds is calculated as;

average velocity = total displacement / total time

average velocity = (2.5 m - 0 m ) / ( 15 s - 0 s ) = 0.167 m/s

(c) The position of the object at 5.0 seconds is calculated as follows;

at 5.0 seconds, the position of the object is traced from the graph as 0.5 m.

(d) The motion of the object between 30 and 40 seconds is calculated as;

velocity = ( 0 m - 4 m ) / ( 40 s - 30 s ) = - 0.4 m/s

Between 30 and 40 seconds, the velocity of the object is decreasing and the object is decelerating.

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The graph that corresponds to 0.1 s in one complete cycle is graph D.

option D is the correct answer.

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years.

Also, the period of a wave is the amount of time it takes for a wave to complete one wave cycle or wavelength.

From the given parameter, the coil rotates 10 times in one second. The period of the coil is calculated as;

Period = 1 s / 10

Period = 0.1 s

From the graphs, the only option that has one complete cycle in one second is option D.

Check option D, half cycle is 0.05 s and one full cycle is 0.1 s.

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