The gravitational force exerted by a proton on an electron is 2x1039 times weaker than the electric force that the proton exerts on an electron? True or False? explain.

Final answer:

To calculate the speed of a muon before it decays, the distance the muon travels and the time it takes are used in the formula L = v x t. For a distance of 9.5 cm and a time of 2.20 microseconds, the speed is found to be approximately 43,182 meters per second.

Explanation:

The question is seeking to calculate the speed of a muon before it decays. Assuming the mean lifetime of a muon is 2.20 microseconds (us) and the muon makes a track 9.5 cm long in this time, we can calculate the muon's speed. Let's consider the formula:

L = v x t

where L is the distance traveled, v is the velocity, and t is the time. Given L = 9.5 cm (or 0.095 m) and t = 2.20 x 10-6 s, we plug in the values:

v = L/t

v = 0.095 m / 2.20 x 10-6 s

v = 4.31818 x 104 m/s

The muon's speed was approximately 43,182 m/s.

Answer:

Explanation:

Given

Speed of block at bottom is [tex]v=3.8\ m/s[/tex]

Distance traveled [tex]s=6\ m[/tex]

initial velocity is zero

using equation of motion

[tex]v^2-u^2=2as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](3.8)^2-0=2\times a\times 6[/tex]

[tex]a=1.203\ m/s^2[/tex]

when it is 3 m from top then

[tex]v^2-u^2=2as[/tex]

[tex]v^2-0=2\times 1.203\times 3[/tex]

[tex]v=2.68\ m/s[/tex]      

The problem is a physics question on kinematics, which requires calculating the speed of a block half-way down a frictionless incline using the kinematic equation for constant acceleration.

The student's question involves finding the speed of a block at a certain point as it slides down a frictionless incline. The kinematics of motion on an inclined plane is the focus here. We are given the speed of the block after traveling 6.00 mm and we need to calculate its speed after it has traveled 3.00 m down the incline.

To solve this, we can use the kinematic equation for constant acceleration, which is stated as:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Since the block starts from rest, u = 0. We can find the acceleration by using the given final speed and distance from the first part of the trip (6.00 mm to the bottom). We substitute the acceleration into the kinematic equation again using s as 3.00 m to find the speed at that point.

Answer:

.A. positive, positive.

Explanation:

When we throw a rock upward , it will decelerate due to gravitation . It will have acceleration in downward direction or - ve acceleration in upward direction.

If we define the ground as the origin and upward direction as positive , anything in upward direction will be positive and in downward direction will be negative . For example , in the case described above , acceleration of rock thrown upward is negative because is  in downward direction .

Its position is in upward direction , so its position is positive with respect to ground.

It is going in upward direction  . So velocity too will be positive.

Answer:

a) [tex](x_2,y_2)=(-45.6,58.8)[/tex]

b) [tex]s=74.4097\ m[/tex]

Explanation:

Given:

a)

As we know that average velocity is total displacement per unit time.

Position of x-coordinate of rhino:

[tex]v_x=\frac{x_2}{t_2}[/tex]

[tex]-3.8=\frac{x_2}{12}[/tex]

[tex]x_2=-45.6\ m[/tex]

Position of y-coordinate of rhino:

[tex]v_y=\frac{y_2}{t_2}[/tex]

[tex]4.9=\frac{y_2}{12}[/tex]

[tex]y_2=58.8\ m[/tex]

b)

Now the distance from the origin:

[tex]s=\sqrt{x_2^2+y_2^2}[/tex]

[tex]s=\sqrt{(-45.6)^2+(58.8)^2}[/tex]

[tex]s=74.4097\ m[/tex]

The question concerns vector kinematics in physics. The rhinoceros's coordinates at 12 seconds are (-45.6,58.8), and its distance from the origin is approximately 74.2 meters.

The phenomenon described in your question essentially relates to Vector Kinematics. The rhinoceros is said to have a velocity in the x-direction of -3.8 m/s and in the y-direction of 4.9 m/s. These velocities are constant and last for 12 seconds.

To find the x- and y-coordinates, we can simply multiply the component velocities by time.

For the x-coordinate:
x = velocity_x * time = -3.8 m/s * 12 s = -45.6 m
And for the y-coordinate:
y = velocity_y * time = 4.9 m/s * 12 s = 58.8 m

So, for part (a) of your question, the rhino's coordinates at t2 = 12.0 s are (-45.6, 58.8).

Now, for part (b), to calculate the rhino's distance from the origin, we can use the Pythagorean theorem which is rooted in the geometrical interpretation of vectors. The distance from the origin (r) can be given by:
r = sqrt[x² + y²] = sqrt[(-45.6 m)² + (58.8 m)²] = approximately 74.2 m

Therefore, the rhino is approximately 74.2 m from the origin.

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The seed launched at an angle of 0° will land 0 cm from the plant regardless of the initial speed because in a projectile motion, angle of 0° results in zero horizontal distance traveled.

This question can be solved by using the formulas of projectile motion. In such motion, the horizontal distance traveled by an object (range) can be calculated using the formula:

R = (v² sin(2θ))/g

where 'v' is the speed of the object, 'θ' is the angle of projection, and 'g' is the acceleration due to gravity.

Here, the launch angle is 0° and the speed is 4.6 m/s. Since the sin(2*0°) is 0, whatever the speed would be, the seed will drop straight down, having no horizontal distance traveled.

So the seed will not move horizontally and it will land 0 cm away from the plant (option not given in choices).

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To solve this problem, we can use the kinematic equations of motion. When a projectile is launched at an angle from the horizontal, we can split its initial velocity into horizontal and vertical components. Since the launch angle is 0°, the initial velocity will have only a horizontal component, and the vertical component will be zero.

Given:

Initial speed,

v 0=4.6m/s

Launch angle,

θ=0∘

Initial height,

h=20cm=0.20m

Acceleration due to gravity,

g= 9.8m/s²

First, we find the time

t it takes for the seed to hit the ground:

Since the initial vertical velocity is 0, we can use the equation:

h= 1/2gt²

Substituting the values:

0.20= 1/2×9.8×t²

0.20=4.9t²

Solving for:

t²= 4.9/0.20

t²=0.040816

t²≈0.202s

Now, we can use this time to find the horizontal distance travelled x. Since the launch angle is 0°, the horizontal velocity remains constant:

x=v₀×t

x=4.6×0.202

x≈0.9292m

So, the seed will land approximately 0.9292m away from the plant.

The closest option provided is (b) 93 cm. However, the correct value is 92.92 cm, which would round to 93 cm. Therefore, the correct answer is (b) 93 cm.

Answer:

 y = 2,645 10⁴ / m²

m=80 kg,  y = 4.13 m

Explanation:

We must solve this problem in two parts, one when it is in free fall and another for the collision with the floor

Let's start by analyzing the crash with the floor,

Initial instant When it arrives but if you start to stop

              p₀ = m v

Final moment. When he stopped

              [tex]p_{f}[/tex] = 0

The momentum is related to the moment by

            I = Δp = p_{f} –p₀

            F t = 0 - mv

            v = -F t / m

Let's calculate

             v = -18000 0.040 / m

             v = -720 / m

The sign indicates that the speed goes down

Now we use energy conservation at two points

Lowest point. Just before crashing

               Em₀ = K = ½ m v²

Highest point. From where it began to fall

               Em_{f} = U = m g y

  Energy is conserved in the fall

               Em₀ = Em_{f}

                ½ m v² = m g y

                y = ½ v² / g

               y = ½ (720 / m)² /9.8

               y = 2,645 10⁴ / m²

For an explicit height value, the object's mass must be known, suppose the masses are m = 80 kg

              y = 2,645 10⁴/80²

              y = 4.13 m

Answer:

Interstellar dust

Explanation:

In modern solar system theory of condensation, interstellar dust, which was lacking in nebular theory, would be the essential component.Cosmic dust is dust that exists in outer space or that has fallen to earth, also called extra-terrestrial dust or spatial powder. The majority of the cosmic dust particle sizes range from a few molecules to 0.1 μm.

Answer:

Average force =  2 kg m/s²

Explanation:

Given

mass = 5 kg

initial velocity = 0 m/s

final velocity = 4 m/s

time = 0.1 sec

Find

average force = ?

Formula

Average force = m (final velocity - initial velocity)/t₂- t₁

                        = (5kg)(0 m/s - 4 m/s)/0 - 0.1

                        = 5 kg (- 4 m/s)/(- 0.1 sec)

                         = 2 kg m/s²

Final answer:

The magnitude of the average force exerted on the ball by the wall is 400 N, computed using the impulse-momentum theorem and the ball's change in momentum during the collision.

Explanation:

To calculate the magnitude of the average force exerted on the ball, we need to first determine the change in momentum of the ball and then use the impulse-momentum theorem. The ball changes direction after hitting the wall, so the final velocity is -4 m/s (negative sign indicating the opposite direction) and the initial velocity is 4 m/s. The total change in velocity is thus -4 m/s - 4 m/s = -8 m/s. The change in momentum (impulse) is given by mass × change in velocity, which is 5 kg × -8 m/s = -40 kg·m/s. The impulse experienced by the ball equals the average force exerted on the ball multiplied by the time of contact, so:

Average Force = - Impulse / Time

Therefore, the magnitude of the average force is:

|Average Force| = 40 kg·m/s / 0.1 s = 400 N.

The negative sign indicates that the force is in the opposite direction of the initial motion, but since the question asks for the magnitude, we take the absolute value.

Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion.

[tex]K.E=\frac{1}{2}mv^2[/tex]

m = mass of object

v= velocity of the object

[tex]K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules[/tex]

b) for a 7 kg bowl to have kinetic energy of 27 Joules:

[tex]27J=\frac{1}{2}\times 7kg\times v^2[/tex]

[tex]v^2=7.7[/tex]

[tex]v=2.8m/s[/tex]

Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy

Answer: v = 5.79 * 10^10m/s.

Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.

Mathematically, we have that

qV = 1/2mv²

q= magnitude of an electronic charge = 1.609*10^-16c

V= potential difference = 95v

m = mass of an electronic charge = 9.11* 10^-31kg.

v = velocity of electron.

Let us substitute the parameters, we have that

1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2

1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²

305.71 * 10^-16 = 9.11 * 10^-31 * v²

v² = 305.71 * 10^-16/ 9.11 * 10^-31

v² = 3.355 * 10^21.

v = √3.355 * 10^21

v = 5.79 * 10^10 m/s

Final answer:

Using the principle of energy conservation and the equation KE = qV = ½ mv^2, the velocity of the electrons in a CRT with an accelerating potential of 95.0 V is calculated to be approximately 5.8 x 10^6 m/s.

Explanation:

To determine how fast electrons will be moving when they hit the screen in a Cathode-Ray Tube (CRT) with an accelerating potential of 95.0 V, one can use the principle of energy conservation. The amount of kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is equal to the change in electrical potential energy (PE), which is given by the equation KE = qV, where q is the charge of the electron (1.602 x 10-19 Coulombs) and V is the potential difference.

The formula for kinetic energy is KE = ½ mv^2, with 'm' representing the mass of the electron (9.109 x 10^-31 kg) and 'v' being the velocity of the electron. Setting these equations equal gives us qV = ½ mv^2. Solving for 'v' and plugging in the values for q, V, and m, we can find the velocity of the electrons when they hit the screen.

Calculation:

qV = ½ mv^2

2qV = mv^2

v = √(2qV/m)

v = √(2 * 1.602 x 10^-19 C * 95.0 V / 9.10^9 x 10^-31 kg)

v ≈ 5.8 x 10^6 m/s

Thus, the electrons would be traveling at approximately 5.8 x 10^6 m/s when they hit the screen.

Answer

given,

mass of the block, m = 0.5 Kg

displacement, x = 30 cm = 0.3 m

Spring constant, k = 2 N/m

a) Angular frequency

   [tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]

   [tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]

  [tex]\omega = 2\ rad/s[/tex]

b) Period of oscillation

   [tex]T=\dfrac{2\pi}{\omega}[/tex]

   [tex]T=\dfrac{2\pi}{2}[/tex]

           T = 3.14 s

c) Position at  t = 2

 x = -A cos ω t

A = 0.3   ω = 2 rad/s

x = -0.3 cos (2 x 2)

x = 0.196 m

d) velocity at t= 2

 v = A ω sin ω t

 v = 0.3 x 2 x sin 4

 v = -0.454 m/s

e) acceleration at t= 2

   a = A ω² cos ω t

   a = 0.3 x 2² cos 4

   a = -0.784 m/s²

Answer:

[tex]L=4.19*10^{-6}H[/tex]

Explanation:

The self-inductance of a solenoid is defined as:

[tex]L=\frac{\mu N^2A}{l}[/tex]

Here [tex]\mu_0[/tex] is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

[tex]L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H[/tex]

The self-inductance of the given solenoid would be approximately 1.395x10^-5 H. The calculation involved substituting known variables into the formula for self-inductance and solving.

The self-inductance of a solenoid can be calculated using the formula: L = µ₀n²A/l where L is the self-inductance, µ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area, and l is the length of the solenoid.

In this scenario, the cross-sectional area (A) = 1.00 × 10⁻⁴ m², length of the solenoid (l) = 30.0 cm = 0.3m, and the number of turns (n) = 100, so n (number of turns per unit length) = 100/0.3 = 333.33 turns/m. µ₀, the permeability of free space is a constant which is given as 4π × 10⁻⁷ T m/A. So, substituting these values into the formula will give:

L (self-inductance) = 4π × 10⁻⁷ T m/A * (333.33 /m)² * 1.00 × 10⁻⁴ m² / 0.3 m = 1.395x10^-5 H, (where H indicates henries, the unit for inductance)

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Answer:

       [tex]\large\boxed{\large\boxed{\frac{a_1}{a_2}=1.2}}[/tex]

Explanation:

Newton's second law states the the net force exerted over a body, [tex]F[/tex], is proportional to the product of its mass, [tex]m[/tex] ,   and its acceleration,   [tex]a[/tex] :

     [tex]F=m\times a[/tex]

As a consequence of Newton's third law, action and reaction forces are equal in magnitude and opposite in direction. Hence, 20 N is the same magnitude of the forces over each sibling.

1. Sibling 1:

     [tex]20N=50kg\times a_1[/tex]

2. Sibling 2:

    [tex]20N=60kg\times a_2[/tex]

3. Ratio sibling 1: sibling 2:

                [tex]\frac{20N}{20N} =\frac{50kg\times a_1}{60kg\times a_2}[/tex]

                [tex]\frac{a_1}{a_2}=\frac{60}{50}=1.2[/tex]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)

C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

[tex]\epsilon_0[/tex] is the vacuum permittivity

By taking a sphere centered in the origin,

[tex]\int E dS = E \cdot 4\pi r^2[/tex]

where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

[tex]q=0[/tex]

And therefore,

[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

[tex]r=R=0.18 m[/tex]

The charge contained within the Gaussian sphere is therefore

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

[tex]E=3.2\cdot 10^7 N/C[/tex]

C)

For this part again, we apply Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field in this case is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]

And again, this is radially inward, so according to the sign convention asked in the problem,

[tex]E=1.2\cdot 10^7 N/C[/tex]

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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.

A) Inside the paint layer:

Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.

Gauss's law equation can be written as:

Φ = Q_enclosed / ε₀

The electric field just inside the paint layer (E₁) is given by:

E₁ * A = Q_enclosed / ε₀

E₁ * 4πr² = -29.0 μC / ε₀

E₁ = (-29.0 μC / ε₀) / (4πr²)

The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.

B) Just outside the paint layer:

The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.

So, E₂ = E₁ (since it's radially outward)

C) 6.00 cm outside the surface of the paint layer:

To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.

E₃ = k * Q / r²

where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).

Remember that the electric field due to a charged object decreases with the square of the distance.

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Answer:

Explanation:

Given

velocity of launch [tex]u=50\ m/s[/tex]

Target is [tex]R=65\m\ away[/tex]

Suppose [tex]\theta [/tex] is the launch angle

We know Range of Projectile is

[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]

[tex]65=\frac{50^2\times \sin 2\theta }{9.8}[/tex]

[tex]\sin 2\theta =0.254[/tex]

because [tex]\sin \theta =\sin (180-\theta )[/tex]

so either [tex]2\theta =14.71[/tex]

[tex]\theta =7.35^{\circ}[/tex]

or [tex]180-2\theta =14.71[/tex]

[tex]\theta =82.64^{\circ}[/tex]

Answer:

t= 0.0003 mm

Explanation:

Given that

mass ,m = 230 mg

m  = 0.23 g

Area ,A= 23 x 17 cm²

A= 391  cm²

Density ,ρ = 19.32 g/cm³

Lets take thinness of the sheet =  t cm

We know that

Mass = Density x Volume

m = ρ A t

Now by putting the values in the above equation we get

0.23 = 19.32 x 391 x t

[tex]t=\dfrac{0.23}{19.32\times 391}\ cm[/tex]

t=0.000030 cm

t= 0.0003 mm

That is why the thickness of the sheet will be 0.0003 mm.

Answer:

A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.

Explanation:

A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.

Each day that goes by, the Earth needs to turn a little further for the sun to return to the same place in the sky. And that extra time is about 4 minutes or 240 seconds.

Final answer:

A solar day is longer than a sidereal day by about 4 minutes, due to Earth's orbit around the Sun requiring additional rotation to align the Sun to the same position in the sky. The sidereal day, on the other hand, represents the true rotation period of Earth as it measures the time it takes for a distant star to reappear in the same position in the sky.

Explanation:

A day measured with respect to the Sun differs from a day measured with respect to the stars primarily in its length due to Earth's simultaneous rotation on its axis and revolution around the Sun. The solar day is the period during which Earth rotates on its axis so that the Sun appears at the same position in the sky on consecutive days. It is about 4 minutes longer than a sidereal day, which is the time it takes for a distant star to return to the same position in the sky.

This difference occurs because as Earth rotates, it also moves along its orbital path around the Sun, so it has to rotate a little more for the Sun to be in the same position overhead. The sidereal day is actually the true rotation period of Earth, being 235.9 seconds shorter than 24 hours, while the solar day includes an additional 4 minutes to account for Earth's movement in orbit.

Answer:

Range will become 4 times of initial range

Explanation:

Let the velocity of projection is u

And angle at which projectile is projected is [tex]\Theta[/tex]

And acceleration due to gravity is [tex]g\ m/sec^2[/tex]

So range of projectile is equal to [tex]R=\frac{u^2sin2\Theta }{g}[/tex]........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity [tex]u_{new}=2u[/tex]

So new range will be equal to [tex]R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}[/tex] .....eqn 2

Now dividing eqn 2 by eqn 1

[tex]\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }[/tex]

[tex]R_{new}=4R[/tex]

So if we double the initial launch speed then range will become 4 times

Doubling the initial launch speed while keeping the launch angle fixed quadruples the range of a projectile due to the direct proportionality between the range and the square of the initial speed, considering factors like launch angle, gravity, and air resistance.

If you keep the launch angle fixed, but double the initial launch speed, the range of a projectile will increase. When the initial speed is doubled, the range increases four times because the range is directly proportional to the square of the initial speed. This phenomenon is influenced by factors such as launch angle, gravity, and air resistance.

Answer:

[tex]\theta=7^o[/tex]

Explanation:

Displacement

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

[tex]\vec r_1=<0,2>\ km[/tex]

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

[tex]\vec r_2=<7cos60^o,-7sin60^o>\ km=<3.5\ ,-6.06>\ km[/tex]

Finally, he drives 9.5 km 35° north of east.

[tex]\vec r_3=<9.5cos35^o,9.5sin35^o>\ km=<7.78\ ,\ 5.45>\ km[/tex]

The total displacement is

[tex]\vec r_t=<0,2>\ km+<3.5\ ,-6.06>\ km+<7.78\ ,\ 5.45>\ km[/tex]

[tex]\vec r_t=<11.28,1.39>\ km[/tex]

The direction can be calculated with

[tex]\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232[/tex]

[tex]\boxed{\theta=7^o}[/tex]

Answer:

0.02 N.

Explanation:

Given:

m = 20 g

= 0.02 kg.

vi = 0.35 m/s

vo = 0.25 m/s

t = 0.1 s

Force, F = m * a

Where,

m = mass of the blood

a = acceleration.

Acceleration, a = (vi - vo)/t

Where,

vi = final velocity

vo = initial velocity

t = time.

a = (0.35 - 0.25)/0.1

= 0.1/0.1

= 1 m/s².

F = 0.02 * 1

= 0.02 N.

Answer:

a) 8.58 m/s upward

b) -2.211 m/s downward

Explanation:

Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.

(a)We have the following equation of motion

[tex]s = v_0t + gt^2/2[/tex]

where [tex]v_0[/tex] is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m

[tex]3.5 = v_01.1 - 9.81*1.1^2/2[/tex]

[tex]3.5 = 1.1v_0 - 5.94 [/tex]

[tex]1.1v_0 = 3.5 + 5.94 = 9.44[/tex]

[tex]v_0 = 9.44 / 1.1 = 8.58 m/s[/tex]

So the keys were thrown initially upward with a speed of 8.58 m/s

(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is

[tex]v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s[/tex]

So they keys would have a downward speed of 2.211 m/s

Answer:

f = 200[Hz]; L=1500[km]  

Explanation:

We know that frequency is the reciprocal of the period, therefore.

[tex]f=\frac{1}{T} \\f=1/0.005\\f=200[Hz][/tex]

And the wavelength is determined using the following equation.

[tex]L=\frac{c}{f} \\where:\\c = light speed = 3*10^{8}[m/s]\\ f = frecuency [Hz]\\L = wavelength [m]\\L= 3*10^{8}/ 200\\ L = 1500000[m] = 1500[km][/tex]

Final answer:

The frequency of a carrier wave with a period of 0.005 seconds is calculated to be 200 Hz. The wavelength cannot be determined without additional information on the wave's speed.

Explanation:

The question involves calculating the frequency and wavelength of a carrier wave given its period (T=0.005 seconds). The frequency (f) of a wave is the reciprocal of its period, defined by the equation f = 1/T. Therefore, for a period of 0.005 seconds, the frequency would be f = 1/0.005 Hz, which equals 200 Hz. To determine the wavelength (λ), we would need the speed (v) of the wave, which is typically the speed of light (c) for electromagnetic waves, but since the speed and specific type of wave are not provided, we can't calculate the wavelength directly from the given information here.

Answer:

(b) the point charge is moved outside the sphere

Explanation:

Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.

[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]

According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.

If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.

Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.

Answer:

[tex]2.7\times 10^{12}[/tex]

Explanation:

We are given that

Electric field =[tex]E=152N/C[/tex]

We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.

We know that

F=qE

Charge on an electron,q=[tex]1.6\times 10^{-19}C[/tex]

Using the formula

Upward electric force=[tex]152\times 1.6\times 10^{-19}[/tex] N

Upward electric force=[tex]F_e=2.43\times 10^{-17}[/tex] N

Mass of electron=[tex]m_e=9.1\times 10^{-31} kg[/tex]

[tex]g=9.8m/s^2[/tex]

Gravitational force=[tex]F=mg[/tex]

Using the formula

Gravitational force=[tex]F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N[/tex]

Ratio of Fe to the Fg=[tex]\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}[/tex]

[tex]\frac{F_e}{F_g}=2.7\times 10^{12}[/tex]

Explanation:

Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.

Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.

Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.

As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.

Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.

Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.

Answer:

The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.

Explanation:

Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.

The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.

Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.

The core of the reactor is often made up of a couple hundred assemblies, according to its power level.

Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.

Answer:

Explanation:

Given

Work required [tex]W=1.35\times 10^{-20}\ J[/tex]

Work done to eject the sodium ion from interior of the cell is given by the product of charge and Potential difference between inner and outer surface of the cell.

[tex]W=q\times V[/tex]

Charge on sodium ion [tex]q=1.6\times 10^{19}\ C[/tex]

[tex]V=\frac{W}{q}[/tex]

[tex]V=\frac{1.35\times 10^{-20}}{1.6\times 10^{19}}[/tex]

[tex]V=0.0718\ V[/tex]            

The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

When potential difference is V and charge of ion is q. Then, work required to eject a postive charge from the interior of cell is computed as,

                   Work  = q * V

It is given that, work W = [tex]1.35*10^{-20}J[/tex]

Charge on sodium ion , [tex]q=1.6*10^{-19}C[/tex]

Substituting above values in above relation.

                   [tex]V=\frac{W}{q}\\ \\V=\frac{1.35*10^{-20} }{1.6*10^{-19} }\\ \\V=0.0844Volt[/tex]

Hence, the magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

Learn more about the potential difference here:

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Answer:

[tex]t=11.1s[/tex]

Explanation:

Given data

Jet Speed v=650 m/s=1.89504 mach

height h=4500 m

Speed of the Sound V=343 m/s

To find

time t

Solution

As we know that

[tex]Vs_{Velocity}=d_{distance} /t_{time} \\d=Vs*t[/tex]

where

[tex]Vs=1.895*V[/tex]

Also from trigonometric properties we know that

[tex]tan\alpha =\frac{Perpendicular}{Base}[/tex]

We have use height h as perpendicular and distance d is base

So

[tex]tan\alpha =h/d\\tan\alpha=h*(1/Vs*t)\\t=\frac{h}{tan\alpha } \frac{1}{Vs}\\[/tex]

First we need to angle α

Since

[tex]Sin\alpha =V/Vs[/tex]

[tex]Sin\alpha =\frac{v}{1.895v}\\\alpha =32^{o}[/tex]

Substitute given values and angle to  find

[tex]t=\frac{h}{tan\alpha } \frac{1}{Vs}\\t=\frac{4500m}{tan(32) } \frac{1}{1.895*343m/s}\\t=11.1s[/tex]

Final answer:

To calculate the time for a shock wave to reach an observer on the ground, divide the altitude of the jet by the speed of sound. For a jet at 4500 m and a sound speed of 343 m/s, it will take approximately 13.12 seconds for the shock wave to reach the ground.

Explanation:

The question pertains to the time it will take for the shock wave from a jet traveling at a constant speed and altitude to reach an observer on the ground.

To calculate this, one has to consider the speed of sound and the altitude of the jet from the ground. Using Pythagoras' theorem, one can determine the distance the sound has to travel to reach the observer. Then divide the distance by the speed of sound to find the time for the shock wave to reach the ground.

Let's denote the speed of sound as v and the altitude of the jet as h. The time t it takes for the shock wave to reach the observer can be calculated using the formula:

t = h / v

As given, the jet is flying at an altitude of 4500 m (4.5 km) and the speed of sound is generally taken to be 343 m/s. Plugging these values into the formula:

t = 4500 m / 343 m/s
t approx 13.12 seconds
Therefore, it will take approximately 13.12 seconds for the shock wave to reach the observer on the ground.

Answer:

53 m/s

Explanation:

Since we take the positive x-direction to be toward the north, an SUV travelling to the south would have a negative velocity, -18m/s, relative to Earth. Velocity of Earth relative to the SUV would be 18m/s. And velocity of police car relative to Earth would be positive, 35 m/s.

Velocity of police relative to SUV would equal to velocity of police car relative to Earth plus velocity of Earth relative to SUV

= 35 + 18 = 53 m/s

Answer: After rounding, the answer to equation A should have  [tex]4.290\times 10^{-19}J[/tex]

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s preceding the first integers are never significant.

All zero’s after the decimal point are always significant.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

Thus for [tex]\frac{6.626\times 10^{-34}Js}{4.630\times 10^{-7}m}\times 2.9979\times 10^8m/s}=4.290299222462\times 10^{-19}J[/tex]

In this problem, 6.626 has 4 significant figures, 4.630 has 4 significant figures and 2.9979 has 5 significant digits thus product will have the least number of significant figures which is 4. So, the answer will be in 4 significant figures. Thus the answer is [tex]4.290\times 10^{-19}J[/tex]

The answer to equation A should have four significant figures, as the least precise measurement in the equation has four significant figures.

When you are calculating with measured values, the rule is that your result cannot be more precise than the least precise measurement. This means when using values in calculations, you need to look at the number of significant figures in the original measurements to determine how many significant figures to use in the final answer.

In the given equation, A: (6.626 x 10−34 J * s) (2.9979 x 108 m / s) / 4.630 x 10−7 m = 4.290299222462 x 10−19 J (unrounded), the least precise measurement is 4.630 x 10−7 m, which has four significant figures. Therefore, after rounding, the answer to equation A should also have four significant figures.

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Answer:

[tex]6.88\times 10^{-6}\ C/m^2[/tex]

Explanation:

Q = Total charge = [tex]6.3\times 10^{-8}\ C[/tex]

[tex]\rho[/tex] = Surface charge density

r = Radius = 2.7 cm

A = Surface area = [tex]4\pi r^2[/tex]

Charge is given by

[tex]Q=A\rho\\\Rightarrow Q=4\pi r^2\times \rho\\\Rightarrow \rho=\dfrac{Q}{4\pi r^2}\\\Rightarrow \rho=\dfrac{6.3\times 10^{-8}}{4\pi (2.7\times 10^{-2})^2}\\\Rightarrow \rho=0.00000687706544224\\\Rightarrow \rho=6.88\times 10^{-6}\ C/m^2[/tex]

The surface charge density is [tex]6.88\times 10^{-6}\ C/m^2[/tex]