Determine the values of mm and nn when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,000,000,000 kgkg. Enter mm and nn, separated by commas.

Answer:

[tex]\frac{E_{A}}{E_{B}}=4[/tex]

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

[tex]E=\frac{F}{q}[/tex].

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

[tex]F=\frac{kQq}{r^{2}}[/tex].

By substitution we get that

[tex]E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}[/tex]

Now, letting [tex]E_{A}[/tex] be the electric field at point A, letting [tex]E_{B}[/tex] be the electric field at point B, and letting R be the distance from the charge to A:

[tex]E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}[/tex].

The ration of the electric fields is

[tex]\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4[/tex]

This means that at half the distance, the electric field is four times stronger.

The ratio of the electric field strength at point A to the electric field strength at point B is 4.

Given to us

A positive charge Q and a point B is twice as far away from Q as point A.

According to Coulomb's law, the electric force between two electrical charges particles is inversely proportional to the square of the distance between them and it is directly proportional to the product of their charges.

[tex]F =\dfrac{k\ Q_1 Q_2}{r^2}[/tex]

We know that an electric field is written as,

[tex]F= EQ[/tex]

Merging the two-equation we get

[tex]E = \dfrac{k Q}{r^2}[/tex]

As we can write

[tex]\dfrac{E_A}{E_B} = \dfrac{\dfrac{k Q}{R^2}}{\dfrac{k Q}{(2R)^2}}[/tex]

[tex]\dfrac{E_A}{E_B} = 4[/tex]

Hence, the ratio of the electric field strength at point A to the electric field strength at point B is 4.

Learn more about Coulomb's Law:

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Explanation:

weight of water displaced=weight of sphere,

Since,  plastic sphere floats in water with 50.0% of its volume submerged

so, water=2 sphere,

so sphere's density = 1/2 of water's  density

Now,  weight of glycerin displaced=weight of sphere,

Also given This same sphere floats in glycerin with 40.0% of its volume submerged.

so glycerin =2.5 sphere,

so sphere's density = .4 of glycerin's = 1/2 of water's.

So glycerin = 1.25 water

Density of glycerin is  5/ 4 times of water density.

Density of sphere is 1/2 times of water density.

Density is defined as ratio of mass to volume , that is density is the amount of mass per unit of volume.

Since,  plastic sphere floats in water with 50% of its volume submerged.

So, Density of sphere = [tex]\frac{1}{2} *water density[/tex]

Since, same sphere floats in glycerin with 40% of its volume submerged.

So, Density of sphere = [tex]\frac{40}{100}*Glycerin density=\frac{2}{5}*glycerindensity[/tex]

Density of Glycerin =  5/2  of sphere density

Density of glycerin =  [tex]\frac{5}{2}*\frac{1}{2}*waterdensity[/tex]

Density of glycerin = [tex]\frac{5}{4} *waterdensity[/tex]

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Answer:

a) The magnitude of the initial velocity is 18.77 m/s.

b) The launching angle is 31.51°.

c) The horizontal component of the velocity at t = 1.900 s is 16.00 m/s.

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) The horizontal component of the position vector at time t = 1.900 s is 30.40 m.

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

Explanation:

Hi there!

The equations for the velocity and position vector of the ball are the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

v = velocity vector at time t.

a and b) First, let´s find the range of the ball, i.e. the horizontal distance traveled by the ball.

The distance traveled by the baseman can be calculated with this equation:

x = v · t

Where:

x =traveled distance.

v = velocity.

t = time

Then:

x = 7.000 m/s · 2.000 s

x = 14.00 m

The baseman runs 14.00 m. Since he was located 18.00 from the home plate, the horizontal distance traveled by the ball is (14.00 m + 18.00 m) 32.00 m.

If we locate the origin of the frame of reference at the point where the ball is hit, the initial vertical and horizontal positions (x0 and y0) are zero. Since the ball is caught at the same height at which it left the bat, the vertical position of the ball when it is caught is 0.

So, the position vector of the ball at the time when it is caught (2 s after it is hit), is the following:

r = (32.00 m, 0 m)

Using the equations of the x- and y-components of the position vector, we can obtain the initial velocity and the angle:

rx = x0 + v0 · t · cos α     (x0 = 0)

ry = y0 + v0 · t · sin α + 1/2 · g · t²         (y0 = 0)

rx = 32.00 m = v0 · 2.000 s · cos α

ry = 0 m = v0 · 2.000 s · sin α - 1/2 · 9.807 m/s² · (2.000 s)²

Solving the first equation for v0:

16.00 m/s / cos α = v0

And replacing v0 in the second equation:

0 m = 32 m · sin α / cos α - 1/2 · 9.807 m/s² · (2.000 s)²

1/2 · 9.807 m/s² · (2.000 s)² = 32 m · tan α

1/2 · 9.807 m/s² · (2.000 s)² / 32 m = tan α

α = 31.51°

b) The launching angle is 31.51°

The initial velocity will be:

16.00 m/s / cos α = v0

16.00 m/s / cos (31.51°) = v0

v0 = 18.77 m/s

a) The magnitude of the initial velocity is 18.77 m/s.

c) Let´s use the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vy = v0 · sin α + g · t

The horizontal component of the velocity does not depend on time (neglecting air resistance).

Then:

vx = 18.77 m/s · cos (31.51°)

vx = 16.00 m/s

0.100 s before the ball is caught, the horizontal component of the velocity is 16.00 m/s.

Now let´s calculate the vertical component of the velocity:

vy = 18.77 m/s · sin (31.51°) - 9.807 m/s² · 1.900 s

vy = -8.823 m/s

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) Let´s use the same equations we have used in part a).

x = x0 + v0 · t · cos α

x = 18.77 m/s · 1.900 s · cos (31.51°)

x = 30.40 m

The horizontal component of the position vector at time t = 1.900 s is 30.40 m

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 18.77 m/s · 1.900 s · sin (31.51°) - 1/2 · 9.807 m/s² · (1.900 s)²

y = 0.9375 m

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

Answer:

A)    N = W cos θ , B)  fr = W sin θ, C) N = W cos θ

Explanation:

To find the required expressions, let's make a free body diagram of the block, see attached.

In the problem of inclined plane the reference system used the x axis is parallel to the plane and the y axis is perpendicular

In this reference system the only force that we must decompose is the body weight F_w = W) for this we use trigonometry

Note that the angle of the plane is equal to the angle between the axis and the weight, therefore

            sin θ = [tex]W_{x}[/tex] / W

            cos θ = [tex]W_{y}[/tex]  / W

            [tex]W_{x}[/tex]  = W sin θ

            [tex]W_{y}[/tex]  = W cos θ

B) Let us write Newton's second law for each axis, in this case as the block is still the acceleration is zero

X axis

             fr - [tex]W_{x}[/tex]  = 0

             fr = W sin θ

A) Y Axis

             N - [tex]W_{y}[/tex]  = 0

             N = W cos θ

C) N = W cos θ

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge [tex]q_{1}= 3.8\times10^{-6}\ C[/tex]

Second charge [tex]q_{2}=3.2\times10^{-6}\ C[/tex]

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

[tex]E_{1}=E_{2}[/tex]

[tex]\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}[/tex]

[tex]\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}[/tex]

[tex]\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}[/tex]

[tex]x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}[/tex]

Put the value into the formula

[tex]x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}[/tex]

[tex]x=1.69\ m[/tex]

Hence, The distance is 1.69 m.

Answer:

(a) see attachment below

(b) see attachment below

(c) the average velocity in part (a) is a better estimate than that in part (b) because it was calculated for a shorter time interval compared to part (b). The basically definition of instantaneous velocity is that it is the limit of the average velocity as the time interview approaches zero (that is as the time interview becomes smaller and smaller). In other words the shorter the time interview the better the accuracy of the result.

(d) see attachment below.

Explanation:

Check the attachment below for the full solution of this problems.

Thank you for reading this post and I hope it is helpful to you. Thank you.

The average velocities between different time intervals are calculated using displacement and time intervals. The best estimate of the bee's instantaneous velocity is determined by one of the calculated average velocities. To find the displacement during a specific time interval, subtract the initial position from the final position.

(a) The average velocity between 4.8 s and 5.3 s can be calculated by finding the displacement and dividing it by the time interval. The displacement is given by subtracting the initial position from the final position, which results in < -4.2, -1.3, 0 > m. Dividing this displacement by the time interval of 0.5 s gives an average velocity of < -8.4, -2.6, 0 > m/s.

(b) Similarly, between 4.8 s and 5.8 s, the displacement is < 3.7, -3.1, 0 > m and the time interval is 1 s. Dividing the displacement by the time interval gives an average velocity of < 3.7, -3.1, 0 > m/s.

(c) The best estimate of the bee's instantaneous velocity at time 4.8 s would be the average velocity between 4.8 s and 5.3 s, which was calculated in part (a) as < 4.4, -2.6, 0 > m/s.

(d) To find the displacement of the bee during the time interval from 4.8 s to 4.85 s, we can subtract the initial position at 4.8 s from the position at 4.85 s. The initial position is < -3.2, 7.7, 0 > m and the position at 4.85 s is < -1, 6.4, 0 > m. Subtracting these vectors gives a displacement of < 2.2, -1.3, 0 > m.

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Answers:

a) [tex]1.05 rad/s[/tex]

b) [tex]1.38 J[/tex]

Explanation:

a) Final angular velocity :

Before solving this part, we have to stay clear that the angular momentum [tex]L[/tex] is conserved, since we are dealing with circular motion, then:

[tex]L_{o}=L_{f}[/tex]

Hence:

[tex]I_{i} \omega_{i}=I_{f} \omega_{f}[/tex] (1)

Where:

[tex]I_{i}[/tex] is the initial moment of inertia of the system

[tex]\omega_{i}=0.73 rad/s[/tex] is the initial angular velocity

[tex]I_{f}[/tex] is the final moment of inertia of the system

[tex]\omega_{f}[/tex] is the final angular velocity

But first, we have to find [tex]I_{i}[/tex] and [tex]I_{f}[/tex]:

[tex]I_{i}=I_{s}+2mr_{i}^{2}[/tex] (2)

[tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex] (3)

Where:

[tex]I_{s}=8 kgm^{2}[/tex] is the student's moment of inertia

[tex]m=2 kg[/tex] is the mass of each object

[tex]r_{i}=1 m[/tex] is the initial radius

[tex]r_{f}=0.28 m[/tex] is the final radius

Then:

[tex]I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}[/tex] (4)

[tex]I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}[/tex] (5)

Substituting the results of (4) and (5) in (1):

[tex](12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}[/tex] (6)

Finding [tex]\omega_{f}[/tex]:

[tex]\omega_{f}=1.05 rad/s[/tex]  (7) This is the final angular speed

b) Change in kinetic energy:

The rotational kinetic energy is defined as:

[tex]K=\frac{1}{2}I \omega^{2}[/tex] (8)

And the change in kinetic energy is:

[tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{i} \omega_{i}^{2}[/tex] (9)

Since we already calculated these values, we can solve (9):

[tex]\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}[/tex] (10)

Finally:

[tex]\Delta K=1.38 J[/tex] This is the change in kinetic energy

To find the final angular speed of the rotating student who pulls objects closer, we apply the conservation of angular momentum, which allows us to solve for the new angular speed. To calculate the change in kinetic energy, we compare the initial and final kinetic energies of the system, considering the changes in moment of inertia and angular speed.

The question involves the concept of conservation of angular momentum and relates to the change in angular speed when a student on a rotating stool moves two 2 kg objects closer to the axis of rotation. Initially, the objects are 1 m away from the rotation axis and the system rotates with an angular speed of 0.73 rad/sec. The moment of inertia (I) of the student plus the stool is 8 kg·m2, and the student pulls the objects to a radius of 0.28 m.

Using conservation of angular momentum, we can say that the initial angular momentum (Linitial) equals the final angular momentum (Lfinal). The formula for angular momentum is L = I·ω, where ω is the angular speed. Therefore:

Linitial = (I + 2·m·rinitial2)·ωinitial

Lfinal = (I + 2·m·rfinal2)·ωfinal

Plugging in the values:

8 kg·m2 + 2(2 kg)(1 m)2)· 0.73 rad/s = (8 kg·m2 + 2(2 kg)(0.28 m)2)·ωfinal

Solving for ωfinal, we find the final angular speed of the student.

For the change in kinetic energy (KE), we calculate the initial and final kinetic energies using KE = (1/2)Iω2, and then find the difference between them to get the change in kinetic energy.

Answer:

Spring Constant K=9290.9550 N/m

Explanation:

Formula we are going to use is:

[tex]2\pi f=\sqrt{\frac{K}{m}}[/tex]

Where:

f is the frequency

K is the spring constant

m is the mass

Given:

Mass of spring oscillator=m=0.206 kg

Resonance Frequency=f=33.8 Hz

Find:

Spring Constant=K=?

Solution:

From Above formula:

Taking Square on both sides

[tex]4(\pi)^2f^2=\frac{K}{m} \\K=4(\pi)^2f^2*m\\K=4(\pi)^2(33.8)^2*(0.206)\\K=9290.9550 N/m[/tex]

Spring Constant K=9290.9550 N/m

Answer:

15086 mm³/s

Explanation:

Given

dr / dt = 3 mm/s

v = 4/3 π r³

dv / dt = (4 / 3)  π 3r²  dr/dt

= 4πr² x 3

= 12πr²

= 12 x 22/7 x 20²

= 15086 mm³/s

The volume of a sphere with a radius increasing at 3 mm/s will be increasing at a rate of approximately 15072 mm³/s when the diameter is 40 mm. This is calculated using the derivative of the sphere's volume formula.

Given that the radius of a sphere is increasing at a rate of 3 mm/s, we need to find how fast the volume is increasing when the diameter is 40 mm. First, we note that the diameter is 40 mm, so the radius is 20 mm.

The formula for the volume of a sphere is:

V = (4/3)πr³

To find the rate at which the volume is increasing, we take the derivative of the volume with respect to time (t):

dV/dt = 4πr²(dr/dt)

We know from the problem that dr/dt (the rate at which the radius is increasing) is 3 mm/s, and r (the radius at the given moment) is 20 mm. Plugging these values into the derivative formula, we get:

dV/dt = 4π(20)²(3)

dV/dt = 4π(400)(3) = 4800π mm³/s

Evaluating this value numerically and rounding to the nearest whole number, we get:

dV/dt ≈ 15072 mm³/s

Explanation:

I assume you're looking for the average angular velocity:

ω_avg = Δθ / Δt

ω_avg = (125° × π / 180°) / (1.3 s)

ω_avg = 1.6782 rad/s

Answer:

14062.5 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision  = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.

Also

E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

The decrease in  kinetic energy of the system during the collision is 14062.5 kg m/s².

The question involves a collision between two train cars of equal mass moving in the same direction, which falls under the principles of momentum conservation and kinetic energy assessment in physics. First, we calculate the final velocity of the combined train cars using the law of conservation of momentum. Then, we compare the initial and final kinetic energies to find the decrease in kinetic energy due to the collision.

pi = m1v1 + m2v2 = (25000×2.50) + (25000×1) = 87500 kg m/s.

Using momentum conservation, vf = pi / (m1 + m2) = 87500/(25000 + 25000) = 1.75 m/s.

KEi = ½m1v1² + ½m2v2² = ½(25000)(2.5)² + ½(25000)(1)² = 90625 kg m/s²

KEf = ½(m1+m2)vf² = ½(25000+25000)(1.75)² = 76562.5 kg m/s².

The difference in kinetic energy: ΔKE = KEi - KEf = 90625 - 76562.5 = 14062.5 kg m/s².

Answer:

The tension in the string is 87.72 N.

Explanation:

Given that,

Mass of two each boxes = 10 kg

Suppose the inclined plane makes an angle is 60° with the horizontal and the coefficient of kinetic friction between the box and plane is 0.15.

We need to calculate the acceleration

Using balance equation

[tex]mg-T=ma[/tex]

Put the value into the formula

[tex]10g-T=10a[/tex]

[tex]98-T=10a[/tex]....(I)

Again using balance equation

[tex]T-mg\sin\theta-F_{f}=ma[/tex]

[tex]T-mg\sin\theta-\mu mg=ma[/tex]

Put the value into the formula

[tex]T-10g\sin60-\mu10\cos60=10a[/tex]

[tex]T-10\times9.8\dfrac{\sqrt{3}}{2}-0.15\times10\times9.8\cos\times\dfrac{1}{2}=10a[/tex]

[tex]T-84.8+7.35=10a[/tex]

[tex]T-77.45=10a[/tex]...(II)

From equation (I) and (II)

[tex]98-77.45=20a[/tex]

[tex]a=\dfrac{98-77.45}{20}[/tex]

[tex]a=1.0275\ m/s^2[/tex]

Put the value of acceleration in equation (I)

[tex]98-T=10\times1.028[/tex]

[tex]-T=10\times1.028-98[/tex]

[tex]T=87.72\ N[/tex]

Hence, The tension in the string is 87.72 N.

The tension in the string connecting the two 10-kilogram boxes is 98 N.

The tension in the string connecting the two 10-kilogram boxes can be determined using Newton's second law and the equilibrium equation for the x-direction. Based on these equations, the tension in the shorter string is twice the tension in the longer string. Therefore, the tension in the string connecting the two boxes would be:

Tension = weight of the hanging mass

So, the tension in the string would be equal to the weight of each 10-kilogram box which is:

Tension = (10 kg) × (9.8 m/s²) = 98 N

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The ambulance will hear a reflected sound with a frequency of 855Hz, which is calculated using the Doppler Effect.

The subject of this question is the Doppler Effect, which states that the observed frequency of a wave depends on the relative speed of the source and the observer. In this case, the source of the sound is the ambulance and the observer is also the ambulance (after reflection of the sound from the hospital). We can use the formula for Doppler Effect in this case:

F' = F *(v+v0) / (v-vS)

Where:

Putting the given values into this formula, we can calculate:

F' = 798Hz * (343m/s + 0) / (343m/s - 108km/h converted to m/s) = 855Hz

Therefore, the frequency of the reflected sound heard by the ambulance is 855Hz. The correct option is b. 855Hz.

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Answer:

[tex]F=33.25\ N[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

Explanation:

Given:

Now the force on the spring on releasing the compression:

[tex]F=k. \Delta x[/tex]

[tex]F=175\times 0.19[/tex]

[tex]F=33.25\ N[/tex]

Now the acceleration due to this force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{33.25}{0.37}[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

Final answer:

The force exerted on a 0.37 kg object when released from a compressed spring with a constant of 175 N/m is 33.25 N. The acceleration at this instant is 89.86 m/s².

Explanation:

We are given a 0.37 kg object attached to a spring with a spring constant of 175 N/m, compressed by 0.19 m. The force exerted by the spring when it is released can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:

F = -kx

Where:

Plugging in the values we get:

F = -(175 N/m) * (0.19 m)

F = -33.25 N

Since the negative sign indicates the direction of the force is opposite to the direction of displacement, we can say that the magnitude of the force is 33.25 N. To find the acceleration at this instant, we can use Newton's second law of motion:

a = F/m

a = 33.25 N / 0.37 kg

a = 89.86 m/s²

The acceleration of the object when it is released is 89.86 m/s².

Answer:

a,b,d and e are correct.

Explanation:

a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.

b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.

d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.

e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct

Therefore the only incorrect option is c .

In this exercise we have to use our knowledge of physics to identify the best alternative that corresponds to each of the given questions:

the only incorrect alternative is the letter C

So analyzing each alternative, we find that:

a) Change of hotness outside change of speed exist a evidence of an interplay.

As the temperature of a main part of written work will only change if heat strength exist flows fashionable or exhausted the  accompanying encircling.

So, alternative A happen correct.

b) Change of the course outside change of speed happen an evidence of an interaction.

The management changes when the object happen secondary a at right angles to increasing speed acceleration which without any doubt wealth the piece exist communicate or outside force applied.

So, alternative B exist correct.

d) Change of shape or arrangement outside change of speed exist evidence of an interplay.

The shape can change only if outside stress happen used ahead of the atom.

So, alternative D exist correct.

e) Change of similarity outside change of speed happen a evidence of an interplay.

The person's individuality of an object can only change if it communicate accompanying environment.

So, alternative E exist correct

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Answer:

 r = 0.1045 cm

Explanation:

The electric field of a point charge is given by the expression

          E = k q / r²

Let's look for the field created by each load

Q₁ charge

           E₁ = k Q₁ / (r-0)²

           E₁ = - 8.99 10⁹ 45.5 10⁻⁹ / r²

           E₁ = - 409.05 / r²

Q₂ charge

           E₂ = k Q₂ / (r-0.14)₂

           E₂ = - 8.99 10⁹ 5.5 10⁻⁹ / (r-0.14)²

           E₂ = -49.45 / (r-0.14)²

The electric field is a vector quantity, so we must use the sum of vectors to get the total field

           E_total = E₁ + E₂

The burden of proof is always considered positive as the two charges are negative the strength is attractive, consider three regions of interest.

- Left of the two charges in this case the two forces are directed to the right and therefore the field is not canceled

- to the right of the two charges, the forces go to the left and the field is not canceled for any distance

- Between the two charges, in this case each force goes in the opposite direction and there is a point for the field to cancel

             0 = E₁ - E₂

             E₁ = E₂

            -409.05 / r² = - 49.45 /(r-0.14)²

              (r-0.14)² = 49.45 / 409.05 r²

             r² - 2 0.14r + 0.14² = 0.12 r²

             r² (1-0.12) - 0.28 r + 0.0196 = 0

             0.88 r² - 0.28 r + 0.0196 = 0

             r² - 0.318 r + 0.0223 = 0

We solve the second degree equation

            r = [0.318 ±√(0.318² - 4 0.0223)] / 2

            r = [0.318 ± 0.109] 2

            r₁ = 0.209 / 2 = 0.1045 m

            r₂ = 0.2135 m

We see that the result in the area of ​​interest between charges is

             r = 0.1045 cm

- There are zones on the sides of the charge, but the force in these zones has a component that takes the test load to the part between the loads, therefore the field is never canceled

Answer:

a)  e% = 49.0%

, b)  e% = 19.7%

Explanation:

The percentage difference is the absolute uncertainty between the value accepted by 100

               e% = Δx / x_value

a) for this case the correct accepted value is x_value = 0.482 m / s²

               Δx = 0.718 - 0.482

               Δx = 0.236 m / s²

               e% = 0.236 / 0.482 100

               e% = 48.96%

b) In this case the results of the two experiments are real and the correct value is the average

              x_value = (0.482 + 0.718) / 2

              x_value = 0.600 m / s²

              e% = (0.718 - 0.600) / 0.600 100

              e% = 19.7%

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

[tex]W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N[/tex]

Converting to lbf

[tex]98.1\times 0.22481=22.053861\ lbf[/tex]

On Moon

[tex]W=10\times 1.624\\\Rightarrow W=16.24\ N[/tex]

Converting to lbf

[tex]16.24\times 0.22481=3.6509144\ lbf[/tex]

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Answer:

Acceleration will be [tex]5706.77rad/sec^2[/tex]

So option (D) will be correct answer

Explanation:

We have given angular speed of the electrical motor [tex]\omega =2695rpm[/tex]

We have to change this angular speed in rad/sec for further calculation

So [tex]\omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec[/tex]

Armature radius is given r = 7.165 cm = 0.07165 m

We have to find the acceleration of edge of motor

Acceleration is given by [tex]a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2[/tex]

So acceleration will be [tex]5706.77rad/sec^2[/tex]

So option (D) will be correct answer

Final answer:

The acceleration of the edge of the rotor is approximately 572400 m/s^2.

Explanation:

The acceleration of the edge of the rotor can be determined using the formula:

acceleration = radius x (angular velocity)²

In this case, the radius of the armature is given as 7.165 cm, which is equal to 0.07165 m. The angular velocity can be calculated by converting the given rpm (2695.0) to radians per second. One revolution is equal to 2π radians, and one minute is equal to 60 seconds, so:

angular velocity = (2695.0 rpm) x (2π rad/1 rev) x (1 min/60 s)

Once the angular velocity is determined, it can be substituted into the formula for acceleration along with the radius:

acceleration = (0.07165 m) x [(2695.0 rpm) x (2π rad/1 rev) x (1 min/60 s)]²

Mathematically solving this equation will give the value of the acceleration in m/s². After calculating the expression, we find that the acceleration of the edge of the rotor is approximately 572400 m/s².

Answer:

t₂ = t₁ / 5

Explanation:

Rotational kinematics using:  ωf = ωi + αt

Starting from rest and speeding up:

ω₁ = 0 + αt₁  ..  Eq1

Starting from ω₁ and slowing to a stop:

0 = ω₁ - 5αt₂  

Substituting for ω₁ from Eq 1

0 = αt₁ - 5αt₂  

5αt₂ = αt₁  

5t₂ = t₁

t₂ = t₁ / 5

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   (I)

Car

x = x0 + v0*t

x = 52 + 21.5t  (II)

then we can apply I = II

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use I or II, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

To find the time it takes for the motorcycle to catch up with the car, we can use the equation of motion and the given information. The motorcycle starts accelerating at t1, so it covers a distance of 0 m. The distance covered by the car until the motorcycle catches up is equal to the initial distance between them, which is 52.0 m.

To find the time it takes for the motorcycle to catch up with the car, we first need to find the acceleration of the car. Since it is traveling at a constant speed, its acceleration is zero. The motorcycle's acceleration is given as 6.00 m/s². We can use the equation of motion, s = ut + 0.5at², to find the distance each vehicle covers from time t1 to t2. The motorcycle will cover a distance of 0 m, as it starts accelerating at t1. The car will cover a distance of (21.5 m/s)(t2 - t1). Since the motorcycle catches up with the car, the distance covered by the car is equal to the initial distance between them, which is 52.0 m.

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The final velocity of the rock is 15 m/s upwards. The rock's initial velocity is upwards at 5 m/s. The acceleration due to gravity is downwards at 9.8 m/s². Hence the correct option is A.

The initial velocity of the rock is upwards at 5 m/s. This is denoted by u.

The acceleration due to gravity is downwards at 9.8 m/s². This is denoted by a.

The time taken for the rock to reach its final velocity is 2 seconds. This is denoted by t.

The final velocity of the rock is denoted by v.

We can use the following equation to calculate the final velocity of the rock:

v = u + at

Substituting the values of u, a, and t, we get:

= 5 + 9.8 * 2

= 15 m/s

Therefore, the final velocity of the rock is 15 m/s upwards. Hence the correct option is A.

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The rock's speed two seconds after being thrown upward is zero. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down.

The rock's speed two seconds after being thrown upward is zero (D).

When an object is thrown upwards, its speed decreases due to the force of gravity pulling it downward. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down. In this case, since the rock's speed is zero after two seconds, it means it has reached its highest point and is now starting to fall back down.

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Final answer:

The initial acceleration of a proton and an electron released 5.50×10−10 mm apart is calculated using Coulomb's law and Newton's second law. The proton's acceleration is approximately 4.91×10¹⁴ m/s², and the electron's acceleration is approximately 8.99×10¹⁶ m/s².

Explanation:

To find the initial acceleration of a proton and an electron when they are released 5.50×10−10 mm apart, we use Coulomb's law to calculate the force and then apply Newton's second law to find the acceleration. First, convert the distance to meters, so 5.50×10−10 mm is 5.50×10−12 m. Coulomb's law gives us the force as F = k×|q1×q2|/r2, where k is Coulomb's constant (9.00×109 N×m2/C2), q1 and q2 are the charges of the proton and electron (1.60×10−19 C), and r is the separation distance. Substituting the values gives F = 8.20×10−10 N. Then, use Newton's second law, a = F/m, where m is the mass of the particle. For the proton, m = 1.67×10−27 kg, and for the electron, m = 9.11×10−31 kg. Therefore, the initial acceleration of the proton is ap ≈ 4.91×1014 m/s2 and the initial acceleration of the electron is ae ≈ 8.99×1016 m/s2.

The initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].

To calculate the initial acceleration of the proton and electron when they are 5.50×10^(-10) mm apart, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:

[tex]\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \][/tex]

Where:

- F is the force between the charges,

- k is Coulomb's constant [tex](\(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\))[/tex],

- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges (in this case, the charge of a proton and an electron, which are [tex]\(1.6 \times 10^{-19} \, \text{C}\) and \(-1.6 \times 10^{-19} \, \text{C}\)[/tex], respectively),

- r is the distance between the charges.

First, we need to convert the distance from milli meters to meters:

[tex]\[ r = 5.50 \times 10^{-10} \, \text{mm} = 5.50 \times 10^{-13} \, \text{m} \][/tex]

Now, we can calculate the force between the charges using Coulomb's law:

[tex]\[ F = \frac{{8.9875 \times 10^9 \times |(1.6 \times 10^{-19}) \times (-1.6 \times 10^{-19})|}}{{(5.50 \times 10^{-13})^2}} \][/tex]

[tex]\[ F = \frac{{8.9875 \times 10^9 \times 2.56 \times 10^{-38}}}{{3.025 \times 10^{-25}}} \][/tex]

[tex]\[ F \approx 7.61 \times 10^{-8} \, \text{N} \][/tex]

Since force F = ma, we can rearrange to find acceleration [tex]\( a = \frac{F}{m} \)[/tex]. For both the proton and electron, we use their respective masses, which are approximately [tex]\(1.67 \times 10^{-27} \, \text{kg}\) and \(9.11 \times 10^{-31} \, \text{kg}\)[/tex], respectively.

For the proton:

[tex]\[ a_p = \frac{7.61 \times 10^{-8}}{1.67 \times 10^{-27}} \][/tex]

[tex]\[ a_p \approx 4.56 \times 10^{19} \, \text{m/s}^2 \][/tex]

For the electron:

[tex]\[ a_e = \frac{7.61 \times 10^{-8}}{9.11 \times 10^{-31}} \][/tex]

[tex]\[ a_e \approx 8.35 \times 10^{21} \, \text{m/s}^2 \][/tex]

So, the initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].

These values represent the accelerations they experience due to the electrostatic force when they are [tex]5.50\times10^{-10}[/tex] mm apart.

Complete question :- Determine the initial acceleration of a proton and an electron when they are initially [tex]5.50\times10^{-10}[/tex] mm apart, a typical atomic distance. Provide the answer in meters per second squared.

Answer:

147.45 Hz

[tex]\Delta f=f_{Lr}-f_{Se}[/tex]

Explanation:

v = Speed of sound in water = 1482 m/s

[tex]v_w[/tex] = Speed of whale = 4.95 m/s

The difference in frequency is given by

[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]

Frequency of the wave in stationary condition

[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]

Ship's frequency which is reflected back

[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]

[tex]f_{Se}=22\ kHz[/tex]

[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]

[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22\times \dfrac{1482+4.95}{1482-4.95}-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]

The difference in wavelength is 147.45 Hz

The Difference in frequency Δf = 147.4 Hz

Expression for Δf in terms of the relevant frequencies ;  Δf = fL[tex]_{r}[/tex] - fL[tex]_{s}[/tex]

Given data :

Speed of sound in water ( V ) = 1482 m/s

Speed of whale ( Vw ) = 4.95 m/s

Frequency of ship sonar ( f ) = 22.0 kHz

Δf =  [tex]f\frac{v + v_{w} }{v - v_{w} } - f \frac{v}{v-v_{w} }[/tex]  ------ ( 1 )

where :

                   = 22 * [ ( 1482 + 4.95) / ( 1482 - 4.95 ) ]

                   = 22.1474 kHz.

Back to equation ( 1 )

Δf = 22.1474 - 22

   = 0.1474 kHz ≈ 147.4 Hz

Hence we can conclude that the Difference in frequency Δf = 147.4 Hz and

Expression for Δf in terms of the relevant frequencies ;  Δf = fL[tex]_{r}[/tex] - fL[tex]_{s}[/tex]

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Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

[tex]a_o=a\times e^{-\lambda t}[/tex]

[tex]\lambda =\frac{0.693}{t_{1/2}}[/tex]

[tex]a=\frac{a_o}{2^n}[/tex]

where,

a = amount of reactant left after n-half lives and time t

[tex]a_o[/tex] = Initial amount of the reactant.

[tex]\lambda = [/tex] decay constant

[tex]t_{1\2}[/tex] = half life of an isotope

n = number of half lives

We have :

[tex]a_o=100.0 g[/tex]

a = ?

t = 552 days

[tex]t_{1/2}=138 days[/tex]

[tex]a=100.0 g\times e^{-\frac{0.693}{138}\times 552}[/tex]

[tex]a=6.254 g[/tex]

[tex]6.254 g=\frac{100.0 g}{2^n}[/tex]

[tex]2^n=\frac{100.0 g}{6.254 g}[/tex]

n = 4

4 half-lives will occur during this period of time.

By dividing the total time of 552 days by the half-life of Polonium-210 of 138 days, we find that 4 half-lives occur in this period.

The number of half-lives that occur during a given period of time can be found by dividing the total amount of time by the period of one half-life. Given that we are asked to find out the number of half-lives for a 552-day period and the half-life of Polonium-210 is 138 days, we can use this approach. So, the calculation would be 552 days ÷ 138 days/half-life = 4 half-lives. Therefore, 4 half-lives of Polonium-210 would occur in a 552-day period.

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Answer:

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Explanation:

Answer:

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Explanation:

The detailed step by step solution to this problem can be forced below. The electric force on the charge is equal in magnitude to (Eq) front columb's law. The electric force on the charge gives it an acceleration of magnitude a. The force is also equal in magnitude to (ma) from newton's second law.

The acceleration of the charge in the electric field is constant and as a result the equations for constant acceleration motion applies to its motion.

KE = 1/2(MV²)

The complete solution can be found in the attachment below.

Answer:

The acceleration and velocity of a body can be related as

V² = U² + 2aS

Where V = the final velocity of the body in (m/s or ft/s)

U = initial velocity of the body (m/s or ft/s)

a = acceleration of the body in (m/s² or ft/s²)

Distance covered by the during that time interval of acceleration in (m or ft)

Explanation:

This equation is very useful in situations where the time interval of motion is not given.

The equation relates the quantities that are along the same axis (x or y).

That is the velocities (initial and final velocities), acceleration and the distance covered must be on the same axis for it to be used correctly. If some of the parameters are on different axis and used together it will lead to errors.

When dealing with multidimensional problems, care should be taken to treat parameters that are along the same axis together and a vector summation be done later to get the requested quantity. Thank you for reading.

Answer:

3.7

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

r = Radius = [tex]5.1\times 10^{-2}\ m[/tex]

A = Area = [tex]\pi r^2[/tex]

V = Voltage = 1120 V

d = Distance of plate seperation = 0.23 mm

Charge is given by

[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0 A}{d}\times V\\\Rightarrow k=\dfrac{Qd}{\epsilon_0 AV}\\\Rightarrow k=\dfrac{1.3\times 10^{-6}\times 0.23\times 10^{-3}}{8.85\times 10^{-12}\times \pi (5.1\times 10^{-2})^2\times 1120}\\\Rightarrow k=3.69164\\\Rightarrow k=3.7[/tex]

The value of dielectric constant is 3.7

Final answer:

To find the dielectric constant (κ) for a parallel-plate capacitor, given its physical dimensions and charge-potential difference, we calculate the capacitance with and without the dielectric and then solve for κ using the relation between charge, capacitance, and potential difference.

Explanation:

The question involves finding the dielectric constant (κ) of the material placed between the plates of a parallel-plate capacitor given the radius of the plates, the separation between them, the charge on the capacitor, and the potential difference across it. The capacitance of a parallel-plate capacitor is given by C = ε_0 κ A / d, where ε_0 is the permittivity of free space, κ is the dielectric constant, A is the area of the plates, and d is the separation between them. The charge (Q) on a capacitor is related to the capacitance (C) and the potential difference (V) across it by Q = CV. From the given information, we can calculate the area of the plates from their radius, and using the given separation, charge, and potential difference, solve for the dielectric constant, κ.

Given:

Radius of the plates, r = 5.10×10−2 m

Separation between the plates, d = 0.23 mm = 0.23×10−3 m

Charge on the capacitor, Q = 1.3 μC = 1.3×10−6 C

Potential difference, V = 1120 V

We can first calculate the area (A) of the plates using the formula for the area of a circle, A = πr2, substitute the value of A and the given values into the formula for C, and then use the relationship Q = CV to solve for κ.

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

To determine the time it takes for a basketball to roll without slipping down an incline, we use the motion formula as well as the formula for acceleration. With the given values, the time is approximately 1.243 seconds.

To solve this problem, first consider the formula that describes the motion of an object rolling down an incline without slipping. The formula is d = (1/2) * a * t^2, where d is the distance, a is the acceleration, and t is the time. We can rearrange this to solve for t, giving us t = sqrt(2 * d / a).

Next, we need to find the acceleration a. This is given by the formula a = g * sin(theta), where g is the acceleration due to gravity (9.81 m/s^2) and theta is the angle of the incline (39.4 degrees). Plugging these values in, we find that a = 9.81 * sin(39.4) = 6.24 m/s^2.

Finally, we can plug these values into our first formula to find the time it takes for the basketball to roll down the incline. This gives us t = sqrt(2 * 4.8 / 6.24) = 1.243 seconds.

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Answer: I have attached the circuit diagram.

Explanation: NOTE that an independent voltage or current source is denoted in a circle. That is how you identify an independent source.

It simply means that the voltage and current source value does not depend on the value of voltage and current elsewhere in the circuit.

It highly utilised when using Mesh analysis on an electric circuit.

Please I have attached the circuit diagram connected in series.

Check it out!