The creative curriculum model claims to be guided by all of the following researchers except

A. Maria Montessori

B. Jean Piaget

C. Howard Gardner

D. Erik Erikson

Answer:

Number of electrons in the system = 973.

Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]

Explanation:

Assumptions:

Given:

Since, the system is comprised of electrons and protons only, therefore,

[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]

The net charge on the system can be written in terms of charges on electrons and protons as

[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]

Putting the value of (2) in (1), we get,

[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]

It is the number of electrons in the system.

Therefore, the number of protons is given by

[tex]\rm n_p = N-n_e=1610-973=637.[/tex]

The total mass of the system is given by

[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]

Final answer:

The system consists of 336 electrons and 1274 protons, given the net charge. To calculate the total mass, multiply the number of each particle by its respective mass and then sum the results., which gives 5192.362×10−31 kg as answer.

Explanation:

To find the number of electrons in a system with a net charge, we can use the formula:

Given that the net charge of the system is -5.376×10−17 C and the charge of an electron is approximately -1.6022×10−19 C, we can calculate the total number of electrons using the formula:

Number of electrons = (-5.376×10−17 C) / (-1.6022×10−19 C/electron)

Number of electrons = 3.3555×102

Since we can’t have a fraction of an electron, we round to the nearest whole number, which is 336 electrons.

To determine the mass of the system, first, we need to find the number of protons, which would be 1610 - 336 electrons = 1274 protons. Now we multiply the number of protons by the mass of a proton and the number of electrons by the mass of an electron to get the total mass:

Mass of electrons = 336 × 9.11×10−31 kg

Mass of protons = 1274 × 1.673×10−27 kg

Adding these two gives the total mass of the system, which is 5192.362×10−31 kg.

Answer:

on moon he can jump 4.2 m high

Explanation:

given data

gravitational acceleration on moon a(m) = 1/6

jump = 0.7 m

to find out

how high could he jump on the moon

solution

we know gravitational acceleration on earth a =  g = 9.8 m/s²

so on moon am =  [tex]9.8 * \frac{1}{6}[/tex]  = 1.633 m/s²

so if he jump on earth his speed will be for height 0.7 m i s

speed v = [tex]\sqrt{2gh}[/tex]

v = [tex]\sqrt{2(9.8)0.7}[/tex]

v = 3.7 m/s

so if he hump on moon

height will be

height = [tex]\frac{v^2}{2*a(m)}[/tex]  

put here value

height =  [tex]\frac{3.7^2}{2*1.633)}[/tex]  

height = 4.2 m

so on moon he can jump 4.2 m high

Answer:

4213.2 Km/h

Explanation:

Given:

Initial Speed of space vehicle relative to Earth, u = 4150 km/h

Mass of rocket motor = 4m

speed of the rocket motor relative to command module , v'  = 79 Km/h

Mass of the command module = m

Now,

let the speed of command module relative to earth be 'v'

From the conservation of momentum, we have

( 4m + m ) × u = m × v + (4m × (v - v'))

or

5m × 4150 = mv + 4mv - 4mv'

or

20750 = 5v - ( 4 × 79 )

or

20750 = 5v - 316

or

v = 4213.2 Km/h

When a charged rod is brought near a neutral metal sphere, the opposite charges are attracted towards the rod, causing an initial attraction. However, when the sphere touches the rod, the charges redistribute, leading to like charges repelling each other with a stronger force than the attraction between opposite charges, resulting in the sphere being repelled.

When a charged rod is brought near a neutral substance, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction.

Thus, when a positively charged glass rod is brought close to the neutral metal sphere, the opposite charges in the metal sphere are attracted towards the rod. However, when the sphere touches the rod, it suddenly flies away from the rod. This is because the charges in the metal sphere redistribute, and now the like charges repel each other with a stronger force compared to the attraction between the opposite charges.

This repulsion causes the metal sphere to be repelled away from the rod, creating a net repulsive force.

The metal sphere is first attracted to the positively charged glass rod due to induction, and then repelled after touching the rod due to the repulsion between like charges.

When the positively charged glass rod is brought close to the uncharged metal sphere, the sphere experiences electrostatic induction. The electrons in the metal sphere are repelled by the positive charges on the rod and move to the far side of the sphere, leaving the side closest to the rod with a net positive charge. This positive side of the sphere is attracted to the negative charges in the rod (or the electrons in the rod are attracted to the positive side of the sphere), causing the sphere to be drawn toward the rod.

As the sphere gets closer to the rod, the attraction increases until they make contact. When the sphere touches the rod, electrons from the rod flow onto the sphere, as the rod has a surplus of electrons due to its negative induction on the side opposite the sphere. This transfer of electrons results in the sphere acquiring a net negative charge, as it gains more electrons.

Once the sphere has the same negative charge as the side of the rod it touched, the electrostatic force of repulsion between the like charges takes over. The negatively charged sphere is now repelled by the negatively charged side of the glass rod. This repulsion is strong enough to overcome the gravitational force and the nylon thread's tension, causing the sphere to suddenly fly away from the rod.

The frequency of  electromagnetic radiation are:

Part A: 6.1432 × 10¹⁴ Hz

Part B: 5.96× 10¹⁴Hz.

The frequency of electromagnetic radiation can be calculated using the speed of light formula:

[tex]v=\frac{c}{\lambda}[/tex]

Where: v is the frequency in hertz (Hz)  

c is the speed of light in a vacuum (299,792,458m/s)

λ is the wavelength in meters (m)

Given the wavelengths in nanometers (nm), we need to convert them to meters by dividing by 10⁹ (since 1 nm = 10⁻⁹m).

Part A: Wavelength = 488.0 nm

λ₁ = 488.0/10⁹

=4.88 × 10⁻⁷

v₁ = c/λ₁

=299,792,458/4.88 × 10⁻⁷

= 61432880.7377× 10⁷

=6.1432 × 10¹⁴

Part B:

Wavelength = 503.0 nm

λ₂ = 503.0/10⁹

=5.03 × 10⁻⁷m

v₂=c/λ₂

=299,792,458/5.03 × 10⁻⁷

=59600886.2× 10⁷

=5.96× 10¹⁴

Hence, the frequency of electromagnetic radiation is 5.96× 10¹⁴Hz.

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The frequency of a wave with a wavelength of 488.0 nm (argon laser) is approximately 6.1475 x 10¹⁴ Hz. The frequency of a wave with a wavelength of 503.0 nm (maximum solar radiation) is approximately 5.9682 x 10¹⁴ Hz.

The wavelength and frequency of electromagnetic waves are inversely related by the speed of light equation, c = λ ν, where c = 3×10⁸ m/s (speed of light), λ = wavelength, and ν = frequency. To find the frequency of a wave when the wavelength is given, rearrange the equation to ν = c/λ. Applying the equation we get:

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Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s[/tex]

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

The stuntwoman will be in the air for approximately 0.677 seconds, and the horizontal distance or freefall between the saddle and the limb when the woman makes her move must be approximately 8.668 meters.

To find out how long the stuntwoman is in the air, we need to apply the physics concept of free fall. The equation for the time of fall under gravity is sqrt(2h/g), where h is the height and g is the acceleration due to gravity. If we substitute the given values, we get √((2 × 2.23m)/9.8m/s²) = 0.677s.

Next, we need to find horizontal distance between the saddle and the limb when the woman makes her move. Since horizontal distance is simply speed × time, and the speed of the horse is constant, we find that 0.677s × 12.8m/s = 8.668m. So, the branch from which the stunt woman drops must be 8.668m in front of the horse when she drops.

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Answer:

15nC & 10 nC

Explanation:

We will use the formula:

[tex]q_{A}q_{B} = \frac{Fr^{2}}{k}[/tex]

Plugging values for F,r, and k, we get:

[tex]\frac{(5.4*10^{-4} N)(0.050m)^{2}}{9.0 * 10x^{9}N*m^{2}/C^{2}}=1.5 * 10 ^{-16} C^{2}[/tex]

Now, use the equation qB = 25nC - qA: (we know this from the problem)

[tex]q_{A}(25 nC - q_{A})=1.5 *10^{-16} C^{2}[/tex]

This is a quadratic equation that is solved to yield

qA = 10nC   or    qA = 15nC.

qB is of course the one that qA is not, but we do not know which is which, however that is irrelevant for the problem.

The charges have the same magnitude after sharing but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.

Given:

Charge on bead A, [tex]qa = 25\ nC = 25 \times 10^{-9}\ C[/tex]

Electric force between the beads, [tex]F = 5.4 \times 10^{-4}\ N[/tex]

Distance between the beads, [tex]r = 5.0\ cm = 0.05\ m[/tex]

Coulomb's constant, [tex]k = 8.99 \times 10^9\ N m^2/C^2[/tex]

Using Coulomb's law, we have on substituting the value:

[tex]F = k \times |q_a \times q_b| / r^2[/tex]

Substitute the values:

[tex]5.4 \times 10^{-4} = (8.99 \times 10^9) \times |(25 \times 10^{-9}) \times q_b| / (0.05)^2[/tex]

Now solve for qb:

[tex]|q_b| = (5.4 \times 10^{-4} \times (0.05)^2) / (8.99 \times 10^9 \times 25 \times 10^{-9})\\|q_b| = 0.0012\ C[/tex]

Since the charges have the same magnitude after sharing, but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.

To summarize:

Charge on bead A (qa) = 0.0012 C

Charge on bead B (qb) = 0.0012 C

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Answer:

a)[tex]E=2.88*10^{-13}N/C[/tex]

b)[tex]E=1.44*10^{-13}N/C[/tex]

c)[tex]F=4.61*10^{-32}N[/tex]

Explanation:

The definition of a electric field produced by a point charge is:

[tex]E=k*q/r^2[/tex]

a)Electric Field due to the alpha particle:

[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C[/tex]

b)Electric Field  due to the electron:

[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C[/tex]

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force magnitude but with opposite direction:

[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N[/tex]

Answer:

216.97 X 10⁻⁵ N

Explanation:

Charge q₁ and q₂ will attract q₃ with force F₁ and F₂ .F₁ and F₂ will be calculated as follows

F₁ = [tex]\frac{9\times10^9\times8\times7.7\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]

F₁ = 82.01 X 10⁻⁵ N

F₂= [tex]\frac{9\times10^9\times8\times15.4\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]

F₂ = 164.02 X 10⁻⁵ N

F₁ and F₂ will act at 60 degree so their resultant will be calculated as follows

R² = (82.01 X 10⁻⁵)² +( 164.02 X 10⁻⁵ )² + 2 X 82.01 X 164.02 X 10⁻¹⁰ Cos 60

R² = 47079.48 X 10⁻¹⁰

R = 216.97 X 10⁻⁵ N

The work done on a particle by the force in the +x-direction from x = x0 to infinity can be calculated using the formula Work = ∫(b/x^n) dx. The limits of integration for the integral are from x0 to infinity.

To calculate the work done by the force in the +x-direction, we can use the formula:

Work = ∫F dx = ∫(b/x^n) dx

Since the force is in the +x-direction and the particle moves along the x-axis from x = x0 to infinity, the limits of integration for the integral are from x0 to infinity.

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Answer:

the mass of the cart is 150 kg

Explanation:

given,

mass of boy(m) = 50 kg

speed of boy (v)= 10 m/s                  

initial velocity of cart (u) = 0                    

final velocity of cart(V) = 2.5 m/s              

mass of the cart(M) = ?                              

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M =[tex]\dfrac{500}{2.5} - 50[/tex]

M = 150 Kg                                          

hence, the mass of the cart is 150 kg

Answer:

Mass of the cart is 750 kg

Given:

Mass of the boy, m = 50 kg

Speed of the boy, v = 10.0 m/s

Final speed of the boy with the cart, v' = 2.5 m/s

Solution:

Initially the cart is at rest and since its on the ground, height, h = 0

Now, by the conservation of energy, mechanical energy before and after will remain conserved:

KE + PE = KE' + PE'          (1)

where

KE = Initial Kinetic energy

KE' = Final Kinetic Energy

PE = Initial Potential Energy

PE' = Final Potential Energy

We know that:

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

Potential energy = mgh

Since, potential energy will remain zero, thus we apply the conservation of Kinetic Energy only.

Let the mass of cart be M, thus the mass of the system, m' = 50 + M

Using eqn (1):

[tex]\frac{1}{2}mv^{2} = \frac{1}{2}m'v^{2}[/tex]

[tex]\frac{1}{2}\times 50\times 10^{2} = \frac{1}{2}(50 + M)\times 2.5^{2}[/tex]

[tex]5000 = 6.25(50 + M)[/tex]

M = 750 kg

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

The force on a charged particle in a magnetic field is calculated using the Lorentz force equation. For the given particle with charge -1.24 x 10⁻⁹ C and velocity vector, the force is determined by the cross product of the velocity and the magnetic field, taking the charge into account.

A particle with a charge of -1.24 x 10⁻⁹ C is moving with an instantaneous velocity of (4.19 X 10⁴ m/s)î + (-3.85 X 10⁴ m/s)ᴇ. To find the force exerted on this particle by a magnetic field we use the Lorentz force equation, which is F = q(v x B), where F is the force on the particle, q is the charge, v is the velocity of the particle, and B is the magnetic field.

For part (a) where the magnetic field B is (1.40 T)î, the velocity vector v is perpendicular to B since v has no i-component, thus the force can be found simply by calculating the magnitude as q*v*B since sin(θ) is 1 for θ = 90°. The direction of the force is given by the right hand rule, considering that both v and B are vectors and the charged particle is negatively charged.

For part (b) where the magnetic field B is (1.40 T)ᴅ, the velocity vector v has no k-component, thus v is perpendicular to B and the same principle applies.

The magnitude of the force in both scenarios is computed using the charge, the magnitude of velocity (which is the combination of both the i and j components), and the magnitude of the magnetic field. The direction will differ based on the cross product between v and B.

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Answer:

A. 30 kg

Explanation:

As we know that,

[tex]1 litre=1000cm^{3}[/tex]

And 1 litre is equivalent to 1kg.

Given that, The volume of the rectangular container is,

[tex]V=20cm\times 30cm\times 50cm\\V=30000cm^{3}[/tex]

And this volume will be equal to [tex]V=30000cm^{3}=30litres[/tex]

And this litres in kg will be equal to,

[tex]V=30litres=30 kg[/tex]

Therefore the mass of this volume of water is 30 kg.

Answer:

(a) 96 ft/s

(b) - 3072 ft/s^2

(c) 0.03125 s

Explanation:

h = 144 ft

u = 0 ft/s

g = 32 ft/s^2

(a) let she strikes the box with velocity v.

Use third equation of motion

[tex]v^{2}=u^{2}+2gh[/tex]

[tex]v^{2}=0^{2}+2\times 32 \times 144[/tex]

v = 96 ft/s

(b) Let the average acceleration is a.

initial velocity, u = 96 ft/s

final velocity, v = 0

h = 18 in = 1.5 ft

Use third equation of motion

[tex]v^{2}=u^{2}+2ah[/tex]

[tex]0^{2}=96^{2}+2\times a \times 1.5[/tex]

a = - 3072 ft/s^2

(c) Let the time taken is t.

Use first equation of motion

v = u + at

0 = 96 - 3072 x t

t = 0.03125 second

Using basic equations of motion, we calculated the speed of the woman just before impact with the ventilator as 29.3 m/s, her average acceleration during the impact as 930 m/s², and the time it took to crush the box as approximately 0.0315 seconds.

The subject of this question is physics, specifically dealing with concepts of kinematics and mechanics. The falling woman problem requires application of the laws of free fall in physics along with some basic algebraic manipulation, and is quite suitable for a high school student.

(a) The speed of the woman just before she collided with the ventilator can be calculated using the formula v=√(2g h) where g is the gravity (9.8 m/s²) and h is the height (144 ft, which is approximately 43.9 m). Plugging in these values gives us v=√(2*9.8*43.9) ≈ 29.3 m/s.

(b) Knowing the depth the box was crushed (converted to meters) we can use the third equation of motion v² = u² + 2a s (s is displacement) we get a = (v² - u²) / 2s = (29.3 m/s)² / 2*0.46m = 930 m/s² as the average acceleration.

(c) Finally, using v = u + at, as initial speed is 0, we get time t = v/a = 29.3m/s / 930m/s² ≈ 0.0315 seconds.

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Answer:

B) n=5

Explanation:

We call the initial value Xo. We start to double this initial value

X1=2*Xo     n=1

We double again:

X2=2*X1=2*(2Xo)      n=2

X2=2*X1=2^2*Xo      n=2

In general:

Xn=(2^n)*Xo

If we want to reach a value 32 times its initial value:

2^n=32

then: n=5

2^5=2*2*2*2*2=32

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

[tex]R =u\cos\theta t [/tex]

Put the value into the formula

[tex]\dfrac{230}{6} = u\cos\theta[/tex]

[tex]u\cos\theta=38.33[/tex].....(I)

We need to calculate the height

Using vertical component

[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]

Put the value in the equation

[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]

[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]

[tex]u\sin\theta=32.06[/tex].....(II)

Dividing equation (II) and (I)

[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]

[tex]\tan\theta=0.8364[/tex]

[tex]\theta=\tan^{-1}0.8364[/tex]

[tex]\theta=39.90^{\circ}[/tex]

(a). We need to calculate the initial speed

Using equation (I)

[tex]u\cos\theta\times t=38.33[/tex]

Put the value into the formula

[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]

[tex]u=49.96\ m/s[/tex]

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Answer:

q2 = - 8 × [tex]10^{-6}[/tex] C

negative sign because attract together

Explanation:

given data

q1 = +3.2 µC = 3.2 × [tex]10^{-6}[/tex] C

distance r = 0.28 m

force F = 2.9 N

to find out

q2 (magnitude and sign)

solution

we know that here if 2 charge is unlike charge

than there will be electrostatic force of attraction , between them

now we apply coulomb law that is

F = [tex]\frac{q1q2}{4\pi \epsilon *r^{2}}[/tex]     ............1

here  we know  [tex]\frac{1}{4\pi \epsilon}}[/tex] = 9 × [tex]10^{9}[/tex] Nm²/C²

so from equation 1

2.9 = 9 × [tex]10^{9}[/tex] × [tex]\frac{3.2*10^{-6}*q2}{0.28^{2}}[/tex]

q2 = [tex]\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}[/tex]

q2 = - 8 × [tex]10^{-6}[/tex] C

Final answer:

Using Coulomb's Law, it's found that the magnitude of charge q2 is 3.2 µC. Since the force experienced by particle 1 is attractive, q2 must have an opposite sign to q1. Therefore, q2 is -3.2 µC.

Explanation:

To determine the magnitude and sign of charge q2, we can use Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is F = k * |q1 * q2| / r², where F is the force between the charges, k is Coulomb's constant (8.988 × 10⁹ N·m²/C²), q1 and q2 are the charges, and r is the distance between the charges.

Given that q1 = +3.2 µC (+3.2 × 10⁻⁶ C), r = 0.28 m, and F = 2.9 N, we can rearrange the equation to solve for q2:

|q2| = (F * r²) / (k * |q1|).

Substituting the given values into the equation, we get:

|q2| = (2.9 N * (0.28 m)²) / (8.988 × 10⁹ N·m²/C² * 3.2 × 10⁻⁶ C),

|q2| = 3.2 µC or -3.2 µC. Because particle 1 experiences an attractive force, it implies that q2 must have an opposite sign to q1. Thus, q2 = -3.2 µC.

Answer:

You would have to drive in north direction 0.41km

Explanation:

We can solve this with trigonometry, We know that:

[tex]sin(\alpha )=\frac{opposite}{hypotenuse}\\where: \\opposite=north\\hypotenuse=distance[/tex]

[tex]North=sin(6.9^o)*3.4km=0.41km[/tex]

So the distance I will have to drive in north direction is 0.41km

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

[tex]Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}[/tex]

Solving for Q:

[tex]Q = \frac{m*V^{2}*r}{K*q}[/tex]

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC

Answer:

The mass of snow accumulated in the lawn in 1 hour equals 57.024 kilograms

Explanation:

Given

Width of lawn = 16 feet

Length of lawn = 20 feet

Thus total area of lawn equals [tex]16\times 20=320sq.feet[/tex]\

Now it is given that 1 square foot accumulates 1350 new snowflakes each minute thus number of snowflakes accumulated by 320 square feet in 1 minute equals

[tex]320\times 1350=432000[/tex]

Now it is given that average mass of each snowflake is [tex]2.20mg=2.20\times 10^{-3}g=2.20\times 10^{-6}kg[/tex]

Hence the mass accumulated per minute equals [tex]432000\times 2.20\times 10^{-6}=0.9504kg[/tex]

Now since there are 60 minutes in 1 hour thus the mass accumulated in 1 hour equals [tex]0.9504kg\times 60=57.024kg[/tex]

Answer:

the speed of the bird is 2.8 m/s in after 11 seconds.

Explanation:

given,

displacement of bird = 28 m

time taken of the displacement of 28 m = 11 s

distance = speed × time                      

velocity = [tex]\dfrac{displacement}{time}[/tex]

             =[tex]\dfrac{28}{10}[/tex]  

             = 2.8 m/s                                                  

hence, the speed of the bird is 2.8 m/s in after 11 seconds.

Answer:

a.) high amplitude, high frequency

Explanation:

Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.

While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.

High pitch means high frequency and high frequency is perceived to have a shrill sound.

The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.

Hence, the loud shrill whistle will have high frequency and high amplitude.

Answer:

B

Explanation:

I think this is right.

Answer:

Average velocity of the bird is 2.54 m/s.

Explanation:

Given that,

Initial speed of the bird, u = 0

It experiences a displacement of, d = 28 m

Time taken, t = 11 s

We need to find the average velocity of the bird. Let v is the average velocity. Mathematically, it is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{28\ m}{11\ s}[/tex]

v = 2.54 m/s

So, the average velocity of the bird is 2.54 m/s. Hence, this is the required solution.

Final answer:

The average velocity of the bird is calculated by dividing the total displacement of 28 m by the total time of 11 s, resulting in an average velocity of 2.545 m/s.

Explanation:

To find the average velocity of a bird that accelerates from rest and experiences a displacement of 28 m in 11 s, we use the formula for average velocity, which is total displacement divided by total time. Since the bird starts from rest and moves in one direction, its average velocity will be the same as its average speed.

Average velocity = Total displacement / Total time = 28 m / 11 s = 2.545 m/s.

Therefore, the average velocity of the bird is 2.545 m/s.

Answer:

[tex]\partial \theta = 0.003[/tex]

Explanation:

we know that

[tex]sin\theta = \frac{3.8}{146.4}[/tex]

[tex]\theta = sin^{-1} \frac{3.8}{146.4}[/tex]

[tex]\theta = 1.484°[/tex]

[tex]\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians[/tex]

as we see that [tex]sin\theta = \theta[/tex]

relative error[tex] \frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}[/tex]

Where X_1 IS HEIGHT OF ROCK

[tex]X_2[/tex] IS THE HEIGHT OF ROAD

[tex]\partial X[/tex] = uncertainity in measuring  distance

[tex]\partial X = 0.05[/tex]

Putting all value to get uncertainity in angle

[tex]\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}[/tex]

solving for [tex]\partial \theta[/tex] we get

[tex]\partial \theta = 0.003[/tex]

Answer:

Speed of the plane over the ground is 429.33 km/hr due West and 70 km/hr towards North

Solution:

According to the question:

Wind flowing in South direction, [tex]v_{s} = 70.0 km/h[/tex]

Air speed, [tex]v_{a} = 435.0 km/h[/tex]

Now,

The velocity vector is required 70 km/h towards north in order to cancel the wind speed towards south.

Therefore,

The ground speed of the plane is given w.r.t fig 1:

[tex]v_{pg} = \sqrt{v_{a^{2}} - v_{s}^{2}}[/tex]

[tex]v_{pg} = \sqrt{435.0^{2} - 70.0^{2}} = 429.33 km/h[/tex]

Final answer:

The speed of the plane over the ground, taking into consideration a wind blowing south at 70.0 km/h and a plane airspeed of 435.0 km/h, is calculated to be approximately 438.3 km/h due west.

Explanation:

An airplane pilot wishes to fly due west, with a wind of 70.0 km/h blowing toward the south, and the plane's airspeed is 435.0 km/h. To determine the speed of the plane over the ground, we must account for both the plane's airspeed and the wind's effect.

The plane's airspeed vector (435.0 km/h) is combined with the wind's vector (70.0 km/h south) to calculate the resultant vector, which represents the plane's actual velocity relative to the ground. This calculation is a vector addition problem that requires the use of the Pythagoras theorem or vector components.

The speed of the plane over the ground (groundspeed) can be found by calculating the magnitude of the resultant vector: Groundspeed = √(air speed² + wind speed²) = √(435² + 70²) km/h, which simplifies to approximately 438.3 km/h to the west. This calculation shows the combined effect of the plane's airspeed and the wind, resulting in the plane's groundspeed.

Answer:

option C

Explanation:

given,

force act on west  = 20 lb

force act at 45° east of north = 80 lb

magnitude of force = ?

∑ F y  =  80 cos 45⁰

    F y =  56.57 lb

magnitude of forces in x- direction

∑ F x = -20 + 80 sin 45⁰

        = 36.57 lb

net force

F = [tex]\sqrt{F_x^2+F_y^2}[/tex]

F = [tex]\sqrt{56.57^2+36.57^2}[/tex]

F = 67.36 lb≅ 67 lb

hence, the correct answer is option C

Answer:

The average speed was 10.12 m/s or 22.96 mi/h

Explanation:

The average speed is defined as:

[tex]v = \frac{d}{t}[/tex]   (1)

Where d is the total distance traveled and t is the passed time.

For the special case of Joseph DeLoach, he traveled a total distance of 200 meters in 19.75 seconds. Those values can be introduced in equation 1:

[tex]v = \frac{200 m}{19.75 s}[/tex]

[tex]v = 10.12 m/s[/tex]

That means that Joseph DeLoach traveled a distance of 10.12 meters per second.

To represent the result in miles per hour, it is necessary to know that 1 mile is equivalent to 1609 meters.

[tex]200 m x \frac{1 mi}{1609 m}[/tex] ⇒ 0.124 mi

it is needed to express the given time in units of hour. 1 hour is equivalent to 3600 seconds.

[tex]19.75 s x \frac{1 h}{3600 s}[/tex] ⇒ 0.0054 h

Then, equation 1 is used with the new representation of the values.

[tex]v = \frac{0.124 mi}{0.0054 h}[/tex]

[tex]v = 22.96 mi/h[/tex]

Answer:

19.2 m/s

Explanation:

We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.

Then the initial position is:

X0 = 0

The initial speed is:

V0 = -4 m/s

It is negative because it is speed upwards and the frame of reference is positive downwards.

Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

He will hit the water at X = 18 m, so:

18 = 0 - 4 * t + 1/2 * 9.81 * t^2

4.9 * t^2 - 4 * t - 18 = 0

Solving this equation electronically:

t = 2.37 s

The diver will hit the water 2.37 s after jumping.

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s

Answer:

D. accelerated

Explanation:

According to Newton's first law of motion, an object always remains in its state of rest or in uniform motion until an external unbalanced force is acted upon the object. This means if an external unbalanced force is acted on the object, it will come to a non-uniform motion i.e., an accelerated motion.

Let me first get you clear that uniform motion is determined by a constant velocity and a state of rest is considered by a zero velocity or speed. But, here we have an unbalanced force acting on the object. This means the object will change its velocity and hence, it has an accelerated motion.

Final answer:

The state of an object being acted upon by an unbalanced force is accelerated. This is because an unbalanced force results in a change in the velocity of the object, causing it to accelerate as per Newton's First Law of Motion.

Explanation:

The state of an object being acted upon by an unbalanced force is accelerated. According to Newton's First Law of Motion, an object at rest stays at rest, and an object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced external force. This implies that the presence of an unbalanced force (net force not equal to zero) changes the velocity of an object, causing it to accelerate.

In other words, option D, accelerated, is the correct answer. When an unbalanced force acts on an object, it results in the object's acceleration, which is a change in its velocity over time. This scenario directly contradicts situations where an object is at rest, at zero speed, or moving with a constant velocity, where no net force or balanced forces are acting on the object.

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

[tex]\tan\theta=\dfrac{4}{15}[/tex]

[tex]\tan\theta=0.266[/tex]

[tex]\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}[/tex]

[tex]\cos\theta=0.966[/tex]

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

[tex]x=(u\cos\theta)t[/tex]

[tex]t=\dfrac{x}{u\cos\theta}[/tex]

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

[tex]y=u\sin\theta(t)-\dfrac{1}{2}gt^2[/tex]

[tex]y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2[/tex]

[tex]y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}[/tex]

Put the value into the formula

[tex]y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})[/tex]

[tex]y=3.517\ m[/tex]

We need to calculate the distance between knothole and the paint ball

[tex]d=h-y[/tex]

[tex]d=4-3.517[/tex]

[tex]d=0.483\ m[/tex]

Hence, The distance between knothole and the paint ball is 0.483 m.