The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the magnitude of the momentum? Be sure to give the correct number of significant figures in your answer.

Answer:

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Explanation:

Here the given values corresponds to the density of a substance

The density of a substance in standard SI unit is given as Kg/m³.

Given value = 43.28lbf/in³

also

we know,

1 lbf = 0.45kg

1in = 0.0254 m

thus,

[tex]\frac{1lbf}{1in^3}=\frac{0.45kg}{0.0254^3m^3}=27460.68\frac{kg}{m^3}[/tex]

therefore,

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

To determine the resistance of this wire is equal to 0.4082 Ohms.

Given the following data:

To determine the resistance of this wire:

First of all, we would determine the area of the wire by using this formula:

[tex]A =\frac{\pi d^2}{4} \\\\A =\frac{3.142 \times (5\times 10^{-4})^2}{4}\\\\A = 1.96 \times 10^{-7}\;m^2[/tex]

Now, we can determine the resistance of this wire:

[tex]R=\frac{\rho L}{A} \\\\R=\frac{3.2 \times 10^{-8} \times 2.5}{1.96 \times 10^{-7}}[/tex]

Resistance = 0.4082 Ohms.

Read more resistance here: https://brainly.com/question/25813707

Answer:

Vertically upwards

Explanation:

Fleming's left hand rule: it states taht when we spread our fore finger, middle finger and thumb in such a way that they are mutually perpendicular to each other, theN thumb indicates the direction of force, fore finger indicates the direction of magnetic field and the middle finger indicates the direction of current.

So according to this rule the direction of force is vertically upwards.

Ham como assim pq ta em ingles

Answer:

0.3 N

Explanation:

mass of cork = 0.01 kg, density of cork = 250 kg/m^3

density of water = 1000 kg/m^3, g = 10 m/s^2

Tension in the rope = Buoyant force acting on the cork - Weight of the cork

Buoyant force = volume of cork x density of water x g

                        = mass x density of water x g / density of cork

                       = 0.01 x 1000 x 10 / 250 = 0.4 N

Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N

Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N

Answer:

Tension = 0.3 N

Explanation:

As we know that the cork is inside water

so the buoyancy force on the cork is counter balanced by tension force in string and weight of the block

So the force equation is given as

[tex]F_b = T + mg[/tex]

now we will have

[tex]Volume = \frac{mass}{density}[/tex]

[tex]V = \frac{0.01}{250} = 4 \times 10^{-5} m^3[/tex]

now buoyancy force on the block is given by

[tex]F_b = \rho V g[/tex]

[tex]F_b = 1000(4 \times 10^{-5})(10)[/tex]

[tex]F_b = 0.4 N[/tex]

now by force balance equation

[tex]0.4 = T + 0.01(10)[/tex]

[tex]T = 0.4 - 0.1 = 0.3 N[/tex]

Final answer:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media.

Explanation:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media:

n1 * sin(θ1) = n2 * sin(θ2)

Given that the incident angle is 50.0° and the refracted angle is 27.7°, we can plug in the values to find the index of refraction:

n1 * sin(50.0°) = n2 * sin(27.7°)

The index of refraction of the glass is the ratio of the index of refraction of the glass to the index of refraction of air, which is approximately 1:

n2 / n1 = 1 / sin(27.7°)

Answer:

[tex]E = 2 \times 10^{-3} V[/tex]

Explanation:

As we know that rate of change in flux will induce EMF

So here we can

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we have

[tex]EMF = \pi r^2\frac{dB}{dt}[/tex]

now we also know that induced EMF is given by

[tex]\int E. dL = \pi r^2\frac{dB}[dt}[/tex]

[tex]E (2\pi r) = \pi r^2\frac{dB}{dt}[/tex]

[tex]E = \frac{r}{2}(\frac{dB}{dt})[/tex]

now plug in all values in it

[tex]E = \frac{0.0133}{2}(0.12 t)[/tex]

[tex]E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m[/tex]

Answer:

Explanation: r₂ = 15 x 10⁻³ m ; r = 12 x 10⁻³m ; r₁ = 10 x 10⁻³ m

     E = Q /4π∈r²x ( r³ - r₁ ³) / ( r₂³ - r₁³ )

         = 15 x 10⁻⁶ x 8.99 x 10⁹x( 12³ - 10³ ) / ( 15³- 10³ ) X 12³

          = 23.92 N/C

The electric field is the force acting on a charge located in a region in space. The electric field at a distance r from the center of the shell is 2.8x10⁶N/C

The electric field (E) can be defined as a region in the space that interacts with electric charges or charged bodies through electric forces.

It is a vectorial field in which an electric charge (q) suffers the effect of the electric force (F). In other words, it represents the force that is acting on a charge located in a certain region.

E = F/q

It is expressed in newton/coulomb (N/C)

In the exposed example, we have the following data,

We need to get the value of the electric field at the given distance r.

E = Q/4πεr² = kQ/r²

Since the r value is located between r1 and r2, we need to get the electric field at that intermedia distance, r. This is a new ratio.

The total charge Q is distributed in the whole area of the shell thickness or density, delimited by r1 and r2.

But to get the electric field at the intermedia distance, r, we only need to get the charge q lower than Q and distributed between the r and r1, which represents a volume v lower than the total volume V of the shell.

So to get the q value, we need to consider the charge volumetric density, σ.

σ = Q/V

Where

But to get Q/V, we need to get the total shell volume value, V.

V = 4/3 π r³

V = 4/3 π (r2³ - r1³)

V = 4/3 π (0.15³ - 0.1³)

V = 4/3 π (0.00237)

V = 0.00993m³

Now that we have the total volume, and we know the value of the total charge, we need to get the density of the shell.

σ = Q/V

σ = 15 μC / 0.00993m³

σ = 15x10⁻⁶ C / 0.00993m³

σ = 0.00151 C/m³

Now, we will calculate the volume v between r and r1, which is what we are interested in.

v = 4/3 π r³

v = 4/3 π (r³ - r1³)

v = 4/3 π (0.12³ - 0.1³)

v = 4/3 π (0.0007)

v = 0.00305m³

Finally, using this new v value and the σ value, we can get the charge q distributed  throughout the volume v.

q = vσ

q = 0.00305m³ x 0.00151 C/m³

q = 4.6x10⁻⁶ C

Now we can get the electric field at a distance r from the center of the shell

E = kq/r²

E = ((8.99 × 10⁹ N ∙ m²/C²) x (4.6x10⁻⁶ C) ) / 0.12²m

E = 2.8x10⁶N/C

Hence, the electric field at a distance r is 2.8x10⁶N/C

You can learn more about electric field at

https://brainly.com/question/8971780

Answer:

25.2 cm

Explanation:

K = 1050 N/m

m = 1.95 kg

h = 1.75 m

By the conservation of energy, the potential energy of the block is converted into the potential energy stored in the spring

m g h = 1/2 x k x y^2

Where, y be the distance by which the spring is compressed.

1.95 x 9.8 x 1.75 = 1/2 x 1050 x y^2

33.44 = 525 x y^2

y = 0.252 m

y = 25.2 cm

The spring will compress by approximately 25.3 cm when a 1.95 kg block is dropped from a height of 1.75 m due to the conversion of gravitational potential energy into elastic potential energy.

The student's question is regarding the compression of a spring due to the gravitational potential energy of a mass that was dropped onto it. This problem can be solved by using the conservation of energy principle, where the potential energy of the block (due to its height) is converted into the spring's elastic potential energy when the spring is compressed.

To find the compression distance (x), we use the following steps:

First we calculate the gravitational potential energy:

PE = mgh = (1.95 kg)(9.81 m/s2)(1.75 m) = 33.637875 J

Then we set this equal to the spring's potential energy and solve for x:

33.637875 J = 1/2 (1050 N/m) x2

x2 = (2 * 33.637875 J) / (1050 N/m)

x2 = 0.064067619

x = 0.253 J

Therefore, the spring compresses by approximately 0.253 meters or 25.3 centimeters.

To find the diffusion rate in the second channel, we can use a formula that takes into account the cross-sectional area and length of the channels. Substituting the given values and simplifying the expression will give the diffusion rate in the second channel.

The diffusion rate is determined by several factors, including the concentration difference, the diffusion constant, temperature, and the size of the molecules. In this case, the diffusion rate in the first channel is given as 4.0 x 10^(-11) kg/s. To find the diffusion rate in the second channel, we can use the formula:

diffusion rate = (cross-sectional area of second channel / cross-sectional area of first channel) x (length of second channel / length of first channel) x diffusion rate of first channel

Substituting the given values:

diffusion rate = (0.30 cm^2 / 0.50 cm^2) x (0.10 cm / 0.25 cm) x (4.0 x 10^(-11) kg/s)

Simplifying the expression will give the diffusion rate in the second channel.

https://brainly.com/question/30697046

#SPJ3

The new diffusion rate in a channel with a cross-sectional area of 0.30 cm² and a length of 0.10 cm would be 6.0 x 10⁻¹¹ kg/s.

The diffusion rate for a solute in a solvent-filled channel is directly proportional to the cross-sectional area and inversely proportional to the length of the channel. Given that the initial diffusion rate is 4.0 x 10-11 kg/s in a channel with an area of 0.50 cm² and length 0.25 cm, to calculate the diffusion rate in a new channel with a different area and length, we use the equation:

Diffusion rate (new) = (Area (new) / Area (old)) x (Length (old) / Length (new)) x Diffusion rate (old)

Substituting the given values:

Diffusion rate (new) = (0.30 cm² / 0.50 cm²) x (0.25 cm / 0.10 cm) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = (0.60) x (2.50) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = 6.0 x 10⁻¹¹ kg/s

This calculation shows how the diffusion rate changes when altering the channel's dimensions.

Answer:

[tex]3.0\cdot 10^{-5} N/m^2[/tex]

Explanation:

The intensity of the radiation at the location of the satellite is given by:

[tex]I=\frac{P}{4\pi r^2}[/tex]

where

[tex]P=3.8\cdot 10^{26}W[/tex] is the power of the emitted radiation

[tex]4\pi r^2[/tex] is the area over which the radiation is emitted (the surface of a sphere), with

[tex]r=5.8\cdot 10^{10}m[/tex] being the radius of the orbit

Substituting,

[tex]I=\frac{3.8\cdot 10^{26}}{4\pi (5.8\cdot 10^{10})^2}=8989 W/m^2[/tex]

Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:

[tex]p=\frac{I}{c}[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex]

is the speed of light. Substituting,

[tex]p=\frac{8989}{3.0\cdot 10^8}=3.0\cdot 10^{-5} N/m^2[/tex]

Answer:

7.07 x 10⁸ N/m²

Explanation:

F = Force applied to the wire = 68 N

d = diameter of the wire = 0.7 mm = 0.7 x 10⁻³ m

r = radius of the wire = (0.5) d = (0.5) (0.7 x 10⁻³) = 0.35 x 10⁻³ m

Area of cross-section of wire is given as

A = (0.25) πr²

A = (0.25) (3.14) (0.35 x 10⁻³)²

A = 9.61625 x 10⁻⁸ m²

Tensile stress is given as

[tex]P = \frac{F}{A}[/tex]

[tex]P = \frac{68}{9.61625\times 10^{-8}}[/tex]

P = 7.07 x 10⁸ N/m²

Given:

Applied Force on wire = 68 N

Diameter of wire, d = 0.7 mm = [tex]0.7\times 10^{-3}[/tex] m

Radius of wire, r = [tex]\frac{d}{2}[/tex] = 0.35 mm = [tex]0.35\times 10^{-3}[/tex] m

Formula used:

Stress =  [tex]\frac{Applied Force}{cross-sectional area}[/tex]

Explanation:

Cross-sectional area, A = [tex]\pi r^{2}[/tex]  =  [tex]\pi (0.35\times 10^{-3})^{2}[/tex]

A = [tex]3.84\times 10^{-7} m^{2}[/tex]

Using the formula for stress:

Stress =  [tex]\frac{68}{3.84\times 10^{-7}}[/tex] =  [tex]1.76\times 10^{8}[/tex]

Explanation:

a) The equivalent resistance for resistors in series is the sum:

R = R₁ + R₂ + R₃

R = 10 Ω + 20 Ω + 30 Ω

R = 60 Ω

b) The resistors are in series, so they have the same current.

I = V/R

I = 12 V / 60 Ω

I = 0.2 A

c) The voltage drop can be found with Ohm's law:

V₁ = I R₁ = (0.2 A) (10 Ω) = 2 V

V₂ = I R₂ = (0.2 A) (20 Ω) = 4 V

V₃ = I R₃ = (0.2 A) (30 Ω) = 6 V

d) The power lost in each resistor is current times voltage drop:

P₁ = I V₁ = (0.2 A) (2 V) = 0.4 W

P₂ = I V₂ = (0.2 A) (4 V) = 0.8 W

P₃ = I V₃ = (0.2 A) (6 V) = 1.2 W

(a) The equivalent resistance of the circuit is 60 Ω.

(b) The current in each resistor is 0.2 A.

(c) The voltage across each resistor is 2 V, 4 V, and 6 V respectively.

(d) The power lost in each resistor is 0.4 W, 0.8 W and 1.2 W.

The equivalent resistance of the circuit is determined as follows;

Rt = R₁ + R₂ + R₃

Rt = 10 + 20 + 30

Rt = 60 Ω

The current in the resistors is calculated as follows;

V = IR

I = V/R

I = 12/60

I = 0.2 A

V1 =IR1 = 0.2 x 10 = 2 V

V2 = IR2  = 0.2 x 20 = 4 V

V3 = 0.2 x 30 = 6 V

P1 = I²R1 = (0.2)² x 10 = 0.4 W

P2 = I²R2 = (0.2)² x 20 = 0.8 W

P3 = I²R3 = (0.2)² x 30 = 1.2 W

Learn more about series circuit here:https://brainly.com/question/19865219

Answer:

i = 323 A

Explanation:

Initial flux due to magnetic field from the coil is given as

[tex]\phi = NB.A[/tex]

here we will have

[tex]N = 190 [/tex]

[tex]B = 1.60 T[/tex]

[tex]A = 0.340 m^2[/tex]

now the flux is given as

[tex]\phi_1 = (190)(1.60)(0.340) = 103.36 T m^2[/tex]

finally current in the electromagnet changed to zero

so final flux in the coil is zero

[tex]\phi_2 = 0[/tex]

now we know that rate of change in flux will induce EMF in the coil

so we will have

[tex]EMF = \frac{\phi_1 - \phi_2}{\Delta t}[/tex]

[tex]EMF = \frac{103.36 - 0}{20 \times 10^{-3}}[/tex]

[tex]EMF = 5168 Volts[/tex]

now induced current is given as

[tex]i = \frac{EMF}{R}[/tex]

[tex]i = \frac{5168}{16} = 323 A[/tex]

Answer:

6.03 x 10^-3 C/Kg

Explanation:

E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2

Acceleration on object is a .

Use second equation of  motion.

S = u t + 1/2 a t^2

0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2

0.0598 = 4.64 + 0.02 x a

a = - 229 m/s^2

Now, F = ma = qE

q / m = a / E = 229 / (3.8 x 10000)

q / m = 6.03 x 10^-3 C/Kg

Calculate the charge-to-mass ratio of the object given initial speed, distance, time, and electric field strength.

Given: Electric field magnitude = 3.80 x 10^4 N/C, initial speed = 2.32 m/s, distance traveled = 5.98 cm, time = 0.200 s, acceleration due to gravity = 9.80 m/s^2.

To find: The object's charge-to-mass ratio (q/m).

Calculations: Determine the acceleration using the provided data, then relate the forces acting on the object (gravity and electric field) to find q/m.

Answers:  

a) How high does the rock get?=1.933m

b)How far downrange from the pedestal does the rock land?=21.25m

Explanation:

This situation is a good example of projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being their main equations as follows:  

x-component:  

[tex]x=V_{o}cos\theta t[/tex]   (1)  

[tex]V_{x}=V_{o}cos\theta[/tex]   (2)  

Where:  

[tex]V_{o}=18m/s[/tex] is the rock's initial speed  

[tex]\theta=20\°[/tex] is the angle

[tex]t[/tex] is the time since the rock is propelled until it hits the ground  

y-component:  

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (3)  

[tex]V_{y}=V_{o}sin\theta-gt[/tex]   (4)  

Where:  

[tex]y_{o}=10m[/tex]  is the initial height of the rock

[tex]y=0[/tex]  is the final height of the rock (when it finally hits the ground)  

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin with the anwers:

Here we are talking about the maximun height [tex]y_{max}[/tex] the rock has in its parabolic motion. This is fulfilled when [tex]V_{y}=0[/tex].

Rewritting (4) with this condition:

[tex]0=V_{o}sin\theta-gt[/tex]   (5)  

Isolating [tex]t[/tex]:

[tex]t=\frac{V_{o}sin\theta}{g}[/tex]  (6)  

Substituting (6) in (3):

[tex]y_{max}=y_{o}+V_{o}sin\theta(\frac{V_{o}sin\theta}{g})-\frac{1}{2}g(\frac{V_{o}sin\theta}{g})^{2}[/tex]   (7)  

[tex]y_{max}=\frac{V_{o}^{2}sin^{2}\theta}{2g}[/tex]   (8)  

Solving:

[tex]y_{max}=\frac{(18m/s)^{2}sin^{2}(20\°)}{2(9.8m/s^{2})}[/tex]   (9)  

Then:

[tex]y_{max}=1.933m[/tex]   (10) This is the maximum height the rock has.

Here we are talking about the maximun horizontal distance [tex]x_{max}[/tex] the rock has in its parabolic motion (this is fulfilled when [tex]y=0[/tex]):

[tex]0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (11)  

Isolating [tex]t[/tex] from (11):

[tex]t=\frac{2V_{o}sin\theta}{g}[/tex] (12)  

Substituting (12) in (1):

[tex]x_{max}=V_{o}cos\theta (\frac{2V_{o}sin\theta}{g})[/tex]   (13)

[tex]x_{max}=\frac{V_{o}^{2}(2cos\theta sin\theta)}{g}[/tex]   (14)

Knowing [tex]sin(2\theta)=2cos\theta sin\theta[/tex]:

[tex]x_{max}=\frac{V_{o}^{2}sin2\theta}{g}[/tex]   (15)

Solving:

[tex]x_{max}=\frac{(18m/s)^{2}sin2(20)}{9.8m/s^{2}}[/tex]   (16)

Finally:

[tex]x_{max}=21.25m[/tex]   (17)

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        [tex]\Delta L=\frac{PL}{AE}[/tex]

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       [tex]A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2[/tex]

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       [tex]0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg[/tex]  

Mass of the climber = 69.38 kg

Answer:

3.7 x 10⁷ m/s

Explanation:

ΔV = Potential difference through which the proton moves = 7.15 MV = 7.15 x 10⁶ Volts

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

v = speed of the proton as it reach zero potential region

m = mass of the proton = 1.66 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (1.66 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.15 x 10⁶)

v = 3.7 x 10⁷ m/s

Final answer:

The speed of protons in a Van de Graaff accelerator transitioning from a 7.15 MV potential to zero potential is about 3.70 x 10^7 m/s, option c

Explanation:

Protons being accelerated in a Van de Graaff accelerator from a 7.15 MV potential to zero potential can be analyzed using the principle of conservation of energy.

The kinetic energy gained by the protons equals the initial potential energy they had, leading to the formula: 1/2 mv^2 = qV.

Calculating this, the speed of the protons when they reach the zero potential region is approximately 3.70 x 10^7 m/s (C).

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

The work done by the pump against gravity in transfering 5 m³ of water to a tank 20 m above the lake is 980000 J

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 5 m³

Density of water = 1000 Kg/m³

Mass = Density × Volume

Mass of water = 1000 × 5

Mass of water (m) = 5000 Kg

Height (h) = 20 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done = mgh

Work done = 5000 × 9.8 × 20

Thus, the work done by the pump is 980000 J

Learn more: https://brainly.com/question/16976412

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

[tex]a_{new} = 8a.[/tex]

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

[tex]a = \frac{v^{2}}{R},[/tex]

[tex]a_{new} =\frac{(2v)^{2}}{0.5R},[/tex]

           [tex]= \frac{4v^{2}}{0.5R}[/tex]

          [tex]= \frac{8 v^{2}}{R}[/tex]

[tex]a_{new} = 8a.[/tex]

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

The acceleration acted on the body moving in a closed circular path towards the center of the curve is known as centripetal acceleration. Due to centripetal acceleration, the body is able to move in a closed circular path.

[tex]a_C=\frac{V^{2} }{R}[/tex]

For old orbit

the radius of orbit = R

Speed pf planet = v

[tex]a_C=\frac{V^{2} }{R}[/tex]

For new orbits

new radius = 0.5R

new speed = 2v

[tex]a_C=\frac{(2V)^{2} }{0.5R}[/tex]³

[tex]a_C=\frac{8V^{2} }{R}[/tex]

Hence,

[tex]a_C_{new}=8a_C__{old}[/tex]

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

To learn more about centripetal acceleration refers to the link

https://brainly.com/question/17689540

Answer:

232 J

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = m C (T − T₀)

Given m = 75.1 g, C = 0.14 J/g/°C, T = 53.8°C, and T₀ = 31.7°C:

q = (75.1 g) (0.14 J/g/°C) (53.8°C − 31.7°C)

q = 232 J

Answer:

[tex]Q=232.36J[/tex]

Explanation:

The heat capacity (C) of a physical system depends on the amount of substance of that system. For a system formed by a single homogeneous substance, it is defined as:

[tex]C=mc(1)[/tex]

Here m is the mass of the system and c is the specific heat capacity.

The heat capacity is defined as the ratio between the heat absorbed by the system and the resulting temperature change:

[tex]C=\frac{Q}{\Delta T}(2)[/tex]

We equal (1) and (2) and solve for Q:

[tex]\frac{Q}{\Delta T}=mc\\Q=mc\Delta T\\Q=mc(T_f-T_i)\\Q=75.1g(0.14\frac{J}{g^\circ C})(53.8^\circ C-31.7^\circ C)\\Q=232.36J[/tex]

Answer:

amplitude, a = 3.076 mm

frequency, f = 10.35 Hz

Explanation:

Vmax = 200 mm /s = 0.2 m/s

Amax = 13 m/s

Let f be the frequency and a be the amplitude.

Use the formula of maximum velocity.

Vmax = ω a

0.2 = ω a    ..... (1)

Use the formula of maximum acceleration.

Amax = ω^2 x a

13 = ω^2 a    ..... (2)

Divide equation (2) by (1)

ω = 13 / 0.2 = 65 rad/s

Put in equation (1)

0.2 = 65 x a

a = 3.076 x 10^-3 m

a = 3.076 mm

Let f be the frequency

ω = 2 π f

f = 65 / (2 x 3.14) = 10.35 Hz  

Answer:

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Explanation:

Let the position of the point where magnetic field is to be determined is at distance "r" from the axis of cylinder

so here total charge lying in this region is

[tex]q = \rho(\pi r^2 L)[/tex]

now magnetic field inside the cylinder is given as

[tex]B = \frac{\mu_0 N i}{L}[/tex]

here current is given as

[tex]i = \frac{q\omega}{2\pi}[/tex]

[tex]i = \frac{\rho (\pi r^2 L) \omega}{2\pi}[/tex]

[tex]i = \frac{\rho r^2 L \omega}{2}[/tex]

now magnetic field is given as

[tex]B = \frac{\mu_0 \rho r^2 L \omega}{2L}[/tex]

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Answer:

The function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3) = 39[/tex]

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

[tex]f(x)' = -4*2x + 24=0[/tex]

[tex]-4*2x + 24=0[/tex]

[tex]8x=24[/tex]

[tex]x=3[/tex]

Now find the second derivative of the function and evaluate at x = 3.

If [tex]f (3) ''< 0[/tex] the function has a maximum

If [tex]f (3) '' >0[/tex] the function has a minimum

[tex]f(x)''= 8[/tex]

Note that:

[tex]f(3)''= -8<0[/tex]

the function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39[/tex]

The question requires determining angles for bright and dark fringes in a double-slit experiment using the formula d sin(\theta) = m\lambda. The student should apply this formula, adjusting it for dark fringes, and solve for the angles for each order m, considering the wavelength and slit separation given.

The question relates to Young's double-slit experiment, which demonstrates the wave nature of light through the interference pattern produced when light passes through two closely spaced slits. The angles locating the fringes can be calculated using the formula

d sin(\theta) = m\lambda, where d is the separation between the slits, \theta is the angle of the fringe from the central maximum, m is the order of the fringe (which can be an integer or half-integer value depending on whether it's a bright or dark fringe), and \lambda is the wavelength of the light.

(a) For the dark fringe with m=0, we expect no fringe to appear as m=0 corresponds to the central maximum which is bright, not dark.

(b) For the bright fringe with m=1, we rearrange the formula to \theta = arcsin(m\lambda / d). Substituting the given values, the angle can be calculated.

(c) For the dark fringe with m=1, the condition is changed to d sin(\theta) = (m + 0.5)\lambda, as dark fringes occur at half-wavelength shifts from the bright fringes.

(d) The calculation for the bright fringe with m=2 follows the same procedure as for m=1, using the appropriate value for m.

The mass of water that needs to evaporate from the skin to reduce a body temperature by 0.45°C is calculated using the body's specific heat capacity, body mass and the latent heat of vaporization of water. First, we find the energy required to cool the body by the temperature change and then find the mass of water that would embody that amount of energy.

The question is asking for the calculation of the mass of water that should evaporate from the skin to reduce the body temperature by 0.45°C. To find this, we need to first understand that evaporation is a main method for body cooling, and it involves a considerable amount of energy being taken from the skin as water changes into vapor.

The energy for evaporating water is explained by the equation Qv = m * Lv where Qv stands for heat energy, m represents mass and Lv is the latent heat of vaporization of water. Given that the specific heat capacity of the body is 4.0 J/g °C and the body mass is 95 kg, the amount of energy required to cool the body by 0.45°C is calculated by multiplying these values (body mass in grams * temperature change in °C * specific heat capacity).

After calculating this energy, we get how much heat needs to be removed from the body to achieve the desired temperature reduction. Lastly, to find the mass of water to be evaporated, we use the equation Qv = m * Lv again but rearrange it as m = Qv / Lv (as Lv = 2256 kJ/kg). This gives us the amount of water that needs to evaporate from the body to reduce the temperature by 0.45°C.

https://brainly.com/question/13433948

#SPJ3

Answer:

[tex]v_f = 20 m/s[/tex]

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2[/tex]

now for pure rolling condition we will have

[tex]v = R\omega[/tex]

also we have

[tex]I = mR^2[/tex]

now we will have

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}[/tex]

[tex]KE = mv^2[/tex]

now by work energy theorem we can say

[tex]W = KE_f - KE_i[/tex]

[tex]842 J = mv_f^2 - mv_i^2[/tex]

[tex]842 = 3(v_f^2) - 3\times 11^2[/tex]

now solve for final speed

[tex]v_f = 20 m/s[/tex]

Answer:

Gravitational force = 89.18 x 10^-31 N

Electric force = 2.4 x 10^-17 N

Magnetic force = 7.68 x 10^-17 N

Explanation:

B = 80 micro tesla = 80 x 10^-6 T  north

E = 150 N/C downward

v = 6 x 10^6 m/s east

Gravitational force = m g = 9.1 x 10^-31 x 9.8 = 89.18 x 10^-31 N

Electric force = q E = 1.6 x 10^-19 x 150 = 2.4 x 10^-17 N

Magnetic force = q v B Sin 90 = 1.6 x 10^-19 x 6 x 10^6 x 80 x 10^-6

                          = 7.68 x 10^-17 N

Answer:

[tex]v_a = 2 v_b[/tex]

Explanation:

As we know that the speed of car A and car B is given by

[tex]v_a[/tex] & [tex]v_b[/tex]

now we know that both cars are decelerated by same deceleration and stopped finally

so the distance moved by the car is given by the equations

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0 - v_a^2 = 2(-a) d_a[/tex]

[tex]d_a = \frac{v_a^2}{2a}[/tex]

similarly we have

[tex]d_b = \frac{v_b^2}{2a}[/tex]

now we know that

[tex]d_a = 4 d_b[/tex]

[tex]\frac{v_a^2}{2a} = 4 \frac{v_b^2}{2a}[/tex]

[tex]v_a = 2 v_b[/tex]