what path would an object have to take to have the distance and the displacement to be equal

Answers

Answer 1

Answer:

When an object move in a straight line without moving back.

Explanation:

Distance is covered by an object is the magnitude of length from one position to the another. It is a scalar quantity.

While displacement is the distance covered in a specific direction. Displacement is a vector quantity. It has both magnitude and direction.

If an object move in a straight path without going back, then, the magnitude of distance will be the same with the magnitude of displacement.

Both distance and displacement are measured in the same unit which is metres.

Therefore, an object have to take a straight path without going back to have the distance and the displacement equal.


Related Questions

In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window

Answers

Answer:

Post indicator valve

Explanation:

Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.

If the terminals of a battery with zero internal resistance are connected across two identical resistors in series, the total power delivered by the battery is 8.00 W. If the same battery is connected across the same resistors in parallel, what is the total power delivered by the battery

Answers

Answer:

24W

Explanation:

The series connection has a resistance of 2R

The parallel connection has a resistance of R/2 .. the resistance has decreased by a factor 4

Assuming the battery still provides the same pd .. the current increases by a factor of 4 .. increasing the power output by a factor of 4 also (P = V x A)

Power output = 4 x 8W .. .. So P = 24 W

The atomic number of a nucleus increases during which nuclear reactions?

Answers

Answer:

Option (A) : Nuclear Fusion and Beta Decay (electron emission)

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

Difference between matter and energy

Answers

Answer:

Energy is the strength and vitality required for sustained physical or mental activity.

Matter occupies space and possesses rest mass, especially as distinct from energy.

Hope this helps! (づ ̄3 ̄)づ╭❤~

Rank these electromagnetic waves on the basisof their speed (in vacuum).
Rank from fastest to slowest. To rankitems as equivalent, overlap them.
yellow light
FM radio wave
green light
X-ray
AM radio wave
infrared wave

Answers

Answer:

All electromagnetic waves travel at the same speed in a vacuum

Explanation:

All the wave listed in the question are electromagnetic waves. The speed of electromagnetic waves (collectively called light) in a vacuum is fixed. Its value is 3×10^8 ms^-1. This is a constant for all electromagnetic waves irrespective of their frequency.

Hence for any electromagnetic wave, its speed is 3×10^8 ms^-1, this will be the common velocity of all the electromagnetic waves listed in the question in a vacuum thus we can not rank them according to speed.

White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (λ = 640 nm) appear in first order?

Answers

Answer:

3.67°

Explanation: Given that λ=640nm , m = 1

Considering the slit separation

d = 1cm/1000

= 1.000×10^-3cm

= 1.000×10-5m

We then have

Sinθ = mλ/d

Sinθ= (1×640×10^-9)/1.000×10-5m

Sinθ = 0.064

θ= sin-1 0.064

θ= 3.669°

= 3.67°

An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pipe? The speed of sound is 343 m/s in air.

Answers

Answer:

The  length is  [tex]l = 8.6 \ m[/tex]

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] ,  [tex]f_2 = 240 \ Hz[/tex]

   The speed of sound in the air is  [tex]v_s = 343 \ m/s[/tex]

Generally the frequency of a given harmonic is mathematically represented as

     [tex]f_n = \frac{n v }{2l}[/tex]

Here  n defines  the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    [tex]220 = \frac{n v}{2l}[/tex]

and  

     [tex]240 = \frac{(n+1) v}{2l}[/tex]

So

   [tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]

=>  [tex]\frac{v}{2l} = 20[/tex]

=>   [tex]l = 8.6 \ m[/tex]

A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth

Answers

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the speed of sound is 341 m/s, how fast is the object moving?

Answers

Answer:

The velocity is  [tex]v = 8.743 \ m/s[/tex]

Explanation:

From the question we are told that

    The frequency of the signal sent out  is  [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]

    The frequency of the signal received is  [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]

     The  speed of sound is  [tex]v_s = 341 \ m/s[/tex]

Generally the frequency of the sound received is  mathematically represented as

         [tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]

where v is the velocity of the object

       =>      [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]

       =>      [tex]1.05263 = \frac{341+v }{341-v}[/tex]

       =>   [tex]358.94 - 1.05263v = 341 + v[/tex]

      =>    [tex]17.947 = 2.05263 v[/tex]

      =>    [tex]v = 8.743 \ m/s[/tex]

A pump is to deliver 10, 000 kg/h of toluene at 1140C and 1.1 atm absolute pressure from the Reboiler of a distillation tower to the second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the Reboiler and the pump is 7 kN/m2. The density of toluene is 886 kg/m3. How far above the pump must the liquid be maintained to avoid cavitation

Answers

Answer:

3.4093

Explanation:

NPSHa = hatm + hel + hf +hva

the elevation head is the hel

friction loss head is hf

NPSHa is the head of vapour pressure of fluid

atmospheric pressure head is hatm

log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]

[tex]A, B, C are fixed[/tex]

log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]

= 4.07827 - 1343.943/333.377

=4.07827 - 4.0313009

= 0.0469691

we take the log

p* = 1.114218

we convert this value to get 111421.8

hvap = 111421.8 * 1/776.14 * 1/9.81

= 14.63

hatm = 1.1 *101325/1 * 1/9.81 *1/776.14

=14.64

hf = 7000/1 * 1/776.14 * 1/9.81

= 0.9193

NPSHa = 2.5

hel = 0.9193 + 2.5 + 14.63 - 14.64

hel = 3.4093

The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.

The vapor pressure can be calculated by,

[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]

Where,

[tex]\mathrm {NPSH_A}[/tex] = available NPSH

[tex]p_i[/tex]     = absolute pressure at the inlet = 1.1 atm

[tex]V_i[/tex]     = average velocity at the inlet = 10, 000 kg/h

[tex]\rho[/tex] = fluid density = 886 kg/m3.  

g = acceleration of gravity = 9.8 m/s²

[tex]p_v[/tex] = vapor pressure of the fluid = ?

Put the values in the equation, we get

[tex]p_v = 1.114\ atm[/tex]

Therefore, the vapor pressure of the liquid is 1.114 atm.

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An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14 mm , the eyepiece a focal length of 21 mm .
A) Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer using two significant figures.
B) How tall is the image mentioned in part A? Express your answer using two significant figures.
C) If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens? Express your answer using two significant figures.
D) Under the conditions of part C, find the overall magnification of the microscope. Express your answer using two significant figures.

Answers

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

Putting the values

[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

What is objective lens?

The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

By using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

Putting the values

[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]

v = 210 mm .

B ) Magnification is the ratio of the size of the image to the size of the an object.

[tex]\rm magnification = \dfrac{v} { u}[/tex]

[tex]M= \dfrac{210} { 15}[/tex]

M= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

[tex]= 14 \times \dfrac{ 250} { 21}[/tex]

= 166.67

= 170 ( in two significant figures )

Hence all the answers are:

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

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A convex spherical mirror has a radius of curvature of magnitude36.0cm.

(a) Determine the position of the virtual image and the magnification for object distances of25.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(b) Determine the position of the virtual image and the magnification for object distances of47.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(c) Are the images in parts (a) and (b) upright or inverted?
The image in part (a) is---Select---uprightinverted
The image in part (b) is---Select---uprightinverted

Answers

Answer:

Explanation:

a )

focal length of convex spherical mirror

f = 36/2  cm = 18 cm

object distance u = - 25 cm

mirror formula

[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} + \frac{1}{- 25} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{18}[/tex]

v = 6.28  cm .

It is positive hence the image will be erect / upright  and formed on the back of the mirror.

For object distance of 47 cm

u = - 47 cm

Putting the values in the mirror formula

[tex]\frac{1}{v} + \frac{1}{- 47} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{ 47} + \frac{1}{18}[/tex]

v = 13  cm

It is positive hence the image will be erect / upright and formed on the back of the mirror.

The base unit prefix used for 1,000× is _____. kilo milli centi deka

Answers

Answer:

[tex]\Large \boxed{\sf kilo}[/tex]

Explanation:

kilo is a prefix that means [tex]1000[/tex] of the base unit.

Answer:

kilo is the correct answer

Explanation:

because my exam says sooo....

An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 82.0 V. The circuit has a resistance of 71.0 Ω and an impedance at this frequency of 107 Ω and an impedance at this frequency of 105Ω. What average power is delivered to the circuit by the source?

Answers

Explanation:

Given that,

Frequency of LCR circuit is 120 Hz

RMS voltage, [tex]V_{rms}=82\ V[/tex]

Resistance of circuit, R = 71 Ω

Impedance, Z = 107 Ω

We need to find the average power is delivered to the circuit by the source. Firstly, finding the rms value of current,

[tex]I_{rms}=\dfrac{V_{rms}}{Z}\\\\I_{rms}=\dfrac{82}{105}\\\\I_{rms}=0.78\ A[/tex]

Power is given by :

[tex]P=I_{rms}V_{rms}\cos\phi[/tex]

[tex]\cos\phi = \dfrac{R}{Z}\\\\\cos\phi = \dfrac{71}{105}\\\\\cos\phi =0.676[/tex]

Now, power,

[tex]P=0.78\times 82\times 0.676\\\\P=43.23\ W[/tex]

So, the average power of 43.23 watts is delivered to the circuit by the source.

A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1), the angle of reflection is 29.0∘ and the angle of refraction is 39.0∘. What is the index of refraction n of the substance?

Answers

Answer:

0.7707

Explanation:

From Snell's law,

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°

=> 1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

Explanation:

The index of refraction n of the substance is 0.7707

Snell law:

Here we know that

n(1) * sin θ1 = n(2) * sinθ2

here

n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection should be via the unknown substance that represent the same as the angle of incidence of air.

So,

θ1 = 29°

1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

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Terms to describe the opposition by a material.to being magnetised is

Answers

Answer:

Repulsion

Explanation:

550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do

Answers

Answer:

The work done by the system is 550 J

Explanation:

Given;

heat added to the system, Q = 550 J

Apply the first law of thermodynamics;

ΔU = Q - W

Where;

ΔU is change in internal energy

Q is the heat added to the system

W is the work done by the system

During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.

0 = Q - W

W = Q

W = 550 J

Therefore, the work done by the system is 550 J

The speed of sound through air is 340 m/s. If a person hears the clap of thunder 9.6 s after seeing the bolt of lightning, how far away is the lightning?

Answers

Explanation:

Distance = speed × time

d = (340 m/s) (9.6 s)

d = 3264 m

A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s

Answers

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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A magnetic field near the floor points down and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counter-clockwise?
a. clockwiseb. counter-clockwise c. no curly E

Answers

Answer:

when a magnetic field near the floors points down and is increasing then the electric field curl (a) clockwise.

Explanation:

The magnetic field this is the area that is around a magnet  which there is presence of magnetic force. The Moving electric charges can create magnetic fields.  we say In physics, that the magnetic field is a field that passes through space and which makes a magnetic force move electric charges.

The Non-coulomb electric field curls ; ( B ) counterclockwise

Non-coulomb electric field also known as induced EMF is the Negative time rate of change of a magnetic flux in a closed loop through the loop. Non-coulomb electric field is expressed as ; Fnc = qEnc

Given that the magnetic field points downwards and the value of the electric field ( ε ) is increasing ( i.e.  ε > 0  ) The direction of the non-coulomb electric field will curl in a counter-clockwise direction.

Hence we can conclude that The Non-coulomb electric field curls in a counterclockwise direction.

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A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.

Answers

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Hope this Helps!! Sorry its late

In _____ research, a group of people of one age is compared to a group of people who are another age.

Answers

Answer:

cross-sectional

Explanation:

The full definition of this is ''a research design in which several different age-groups of participants are studied at one particular point in time.''

A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.

Answers

Answer:

5.29×10^-7

Explanation:

shear stress τ = F/ A

shear deformation δ = (VL)/ (AG)

= (τL)/ G

V=shear force

L=height of disk=6.50×10^-2

A=cross sectional area

G= shear modulus= (1.60x10^9N/m^2)

A=πd^2/4

Then substitute the values we have

4×(375N)(0.00750m)

________________ = δ

(π*0.00650^2)(1.60x10^9N/m^2)

= 5.29×10^-7

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

Answers

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]

where [tex]M_{0}[/tex] is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]

where

[tex]\beta = V/c[/tex]

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

the equation reduces to

1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]

square both sides of the equation, we have

3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]

3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1

2.0625 = 3.0625[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 0.67

β = 0.819

substitute for  [tex]\beta = V/c[/tex]

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static friction between tires and road is 0.50?

Answers

Answer:

         v= 21.47m/s      

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

[tex]U*m*g=\frac{mv^2}{r}[/tex]

[tex]U*g=\frac{v^2}{r}[/tex]

substituting our given data in to expression we can solve for the speed V

[tex]0.5*9.81=\frac{v^2}{94}[/tex]

making v the subject of formula we have

[tex]0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47[/tex]

v= 21.47m/s

hence the maximum velocity of the car is 21.47m/s

Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.

Answers

Answer:

A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.

Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found

Answers

Answer:

only in or near star-forming clouds

Explanation:

When in the galactic disk, Red and orange stars are found evenly spread so here Blue stars are hot and therefore massive and therefore short-lived,  that is means they never have time to venture far from the places, where they were born. so correct answer is blue stars are typically found only in or near star-forming clouds

From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:

a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down

Answers

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s

A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?

Answers

Answer:

386.13 N

Explanation:

The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).

Therefore, KE of the ball

[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]

Now, workdone in moving the glove

W= Fd

where F = Force applied, d = displacement of the glove= 0.23 cm.

88.81 = F×0.23

F= 88.81/0.23 = 386.13 N

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