Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s. Sue applies her brakes but can accelerate only at −1.90 m/s2 because the road is wet. Will there be a collision?

Answers

Answer 1

Answer:

Hence, there will be a collision

Explanation:

First we calculate total distance covered by the speedy sue's car before coming to rest:

2as = Vf² - Vi²

where,

a = deceleration = - 1.9 m/s²

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 35 m/s

Therefore,

2(-1.9 m/s²)s = (0 m/s)² - (35 m/s)²

s = 322.37 m

Now, we calculate time taken by car to stop:

Vf = Vi + at

0 m/s = 35 m/s + (-1.9 m/s²)t

t =  18.42 s

Now, we calculate distance traveled by van in this time:

s₁ = V₁t

where,

s₁ = distance traveled by van = ?

V₁ = speed of van = 5.2 m/s

Therefore,

s₁ = (5.2 m/s)(18.42 s)

s₁ = 95.78 m

Now, for collision to occur, the following relation must be satisfied:

s ≥ 160 m + s₁

using values:

322.37 m > 160 m + 95.78 m

322.37 m > 255.78 m

Hence, there will be a collision


Related Questions

A virus has a diameter of 5.2×10−8m. What is the most appropriate prefix for reporting the diameter of the virus?

Answers

Answer:

5.2 centimicro meter or 5.2 cμm

Explanation:

Prefixes allow scientists report numbers logically depending on the order of the magnitude of the numbers. For example to represent 4 x 10⁻³m, it is more handy to say : 4 millimetre, where milli- is the prefix for 10⁻³.

Common prefixes are:

deci = 10⁻¹                     deca = 10¹                    

centi = 10⁻²                   hecto = 10²

milli  = 10⁻³                    kilo  = 10³                      

micro = 10⁻⁶                  mega = 10⁶

nano = 10⁻⁹                   giga = 10⁹

pico = 10⁻¹²                   terra = 10¹²

femto = 10⁻¹⁵                peta = 10¹⁵

Now, let's represent the given diameter: 5.2 x 10⁻⁸ m

Since the 10⁻⁸ prefix is not common (or maybe doesn't even exist), the way around this is to break it into bits such as:

10⁻⁸ = 10⁻² x 10⁻⁶           [10⁻² = centi,  10⁻⁶ = micro ]

10⁻⁸ = centimicro

Therefore, 5.2 x 10⁻⁸ m becomes 5.2 centimicro meter.

The symbol is 5.2 cμm

Where c = centi, μ = micro.

g A person walks 70 m west, then turns and walks 30 m east. Find the average speed and velocity if the time for the total trip is 45 seconds.

Answers

Answer:

Average speed = 2.22m/s

Velocity = 0.89 m/s west

Explanation:

(i) Since speed is a scalar quantity, the direction of movement by the person is irrelevant. Therefore,

Average speed = total distance traveled / time taken

Where;

Total distance = 70m + 30m = 100m

Time taken = 45 seconds

Average speed = 100 / 45 = 2.22m/s

(ii) However, velocity is a vector quantity and so the direction of movement by the person is of utmost importance. Therefore,

Velocity = displacement / time taken

Where;

displacement = -70m + 30m    [West direction is taken as negative and east is taken as positive]

displacement = -40m     [The negative sign just means that the net displacement is in the direction of the West].

Thus, displacement can be written as 40m west.

Therefore,

Velocity = 40 / 45 = 0.89m/s west

The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is

Answers

Answer:

The new period is 16 s

Explanation:

The period of a simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]

Where [tex]T[/tex] is the period

[tex]L[/tex] is the length of the string

[tex]g[/tex] is the acceleration due to gravity

2π is constant

For the first experiment,

[tex]T =8 s[/tex]

[tex]g = gE[/tex]

Then,

[tex]8 = 2\pi \sqrt{\frac{L}{gE} } \\[/tex]

[tex]8\sqrt{gE} = 2\pi \sqrt{L}[/tex] ....... (1)

For the second experiment

[tex]T = T_{2}[/tex]

[tex]g = \frac{1}{4} gE[/tex]

Hence,

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex] becomes

[tex]T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }[/tex]

Then,

[tex]T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}[/tex] ....... (2)

Since 2π is constant and

The same pendulum is used for the second experiment, then [tex]L[/tex] is also constant.

∴ [tex]2\pi \sqrt{L}[/tex] is constant for both experiment.

Hence, we can equate equations (1) and (2) such that

[tex]8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }[/tex]

Then,

[tex]\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s[/tex]

Hence, the new period is 16 s

A simple pendulum swings for 8 seconds. The new period is determined by repeating the pendulum experiment on a plant with g = 1/4 gE.

Move to a position where the gravitational acceleration is greater.

- Reduce the pendulum's length.

Explanation: The frequency of a simple pendulum is given by where L is the length of the pendulum g is the gravitational acceleration

From the formula, we notice that - As the value of g increases, the frequency of the pendulum increases as well

- As the value of L increases, the frequency of the pendulum decreases

So, in order to increase the frequency of the pendulum, we can: - Move to a location where the acceleration due to gravity, g, is larger
To learm more about pendulum's visit: https://brainly.com/question/12495919

What is its inside radius, in centimeters, if it holds 360 g of coffee when filled to a depth of 8.5 cm

Answers

Answer:

1.14*10^-3m

Explanation:

We know that

Volume= mass/ density

But

Area x depth= mass/density

πr²*depth= mass/density

r²=m/πdp

So

r= √0.35/3.142*0.085*1000

r= 1.14*10^-3m

A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking the surface between them at a 60.0°60.0° incident angle. What is the angle between the two colors in water?

Answers

Answer:

0.98°

Explanation:

First we find the refracting angle of orange in water

Which is given as

စr= sin^-1. ( 1.456 sin60°/1.33)

= 71.2°

Then that of violet in water

စv=sin^-1. ( 1.468sin60°/1.342)

= 72.3°

So angle between boths colours is the difference

71.2- 72.3= 0.98°

A 0.450-kg ice puck, moving east with a speed of 5.34 m/s , has a head-on collision with a 0.990-kg puck initially at rest. Assume that the collision is perfectly elastic.1. What is the speed of the 0.450-kg puck after the collision?2. What is the direction of the velocity of the 0.450-kg puck after the collision?3. What ise the speed of the 0.990-kg puck after the collision?4. What is the direction of the velocity of the 0.990-kg puck after the collision?

Answers

Answer:

(1) -2.0025m/s.

(2) It moves west.

(3) 3.3375m/s.

(4) It moves east.

Explanation:

Assuming the collision occurred in an isolated system. This means that total momentum of the system of pucks is conserved. Since no external forces are acting on these pucks, the momentum of the pucks before collision is equal to the momentum of the pucks after collision. i.e

(p₁)₀ + (p₂)₀ = (p₁)₁ + (p₂)₁              -------------(i)

Where;

(p₁)₀ = momentum of the 0.45kg puck before collision

(p₂)₀ = momentum of the 0.990kg puck before collision

(p₁)₁ = momentum of the 0.45kg puck after collision

(p₂)₁ = momentum of the 0.990kg puck after collision

But;

(p₁)₀ = m₁ u₁  

[m₁ = mass of the 0.45kg, u₁ = speed of the 0.45kg before collision]

(p₂)₀ = m₂u₂

[m₂ = mass of the 0.990kg, u₂ = speed of the 0.990kg before collision]

(p₁)₁ = m₁v₁

[m₁ = mass of the 0.45kg, v₁ = speed of the 0.45kg after collision]

(p₂)₁ = m₂v₂

[m₂ = mass of the 0.990kg, v₂ = speed of the 0.990kg after collision]

Equation (i) then becomes;

m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂    ----------------(ii)

From the question:

m₁ = 0.450kg

u₁ = +5.34m/s   [Taking east direction as positive]

m₂ = 0.990kg

u₂ = 0m/s    [since the second puck is initially at rest]

Substitute these values into equation (ii)

(0.450 x 5.34) + (0.990 x 0) = 0.45 v₁ + 0.990 v₂

2.403 + 0 = 0.45 v₁ + 0.990 v₂

2.403 = 0.45 v₁ + 0.990 v₂           ------------------(iii)

Also, since the collision is perfectly elastic, it means that the kinetic energy is conserved. i.e the total kinetic energy before collision is equal to the total kinetic energy after collision.

=> [tex]\frac{1}{2}[/tex]m₁ u²₁ + [tex]\frac{1}{2}[/tex]m₂u²₂ = [tex]\frac{1}{2}[/tex]m₁v²₁ + [tex]\frac{1}{2}[/tex]m₂v²₂

Substitute the necessary values into the above equation:

[[tex]\frac{1}{2}[/tex] x 0.45 x 5.34²] + [0] = [[tex]\frac{1}{2}[/tex] x 0.45 x v²₁] + [[tex]\frac{1}{2}[/tex] x 0.990 x v²₂]

[6.41601] = [0.225 x v²₁] + [0.495 x v²₂]  ------------------(iv)

Now let's solve equations (iii) and (iv) simultaneously

2.403 = 0.45 v₁ + 0.990 v₂

6.41601 = 0.225 x v²₁ + 0.495 x v²₂

let

v₁ = x

v₂ = y

2.403 = 0.45 x + 0.990 y                  ------------(5)

6.41601 = 0.225 x² + 0.495 y²         -------------(6)

From equation (5), make x subject of the formula

2.403 = 0.45x + 0.990y

0.45x = 2.403 - 0.990y         [divide through by 0.45]

x = 5.34 - 2.2y    ----------------(m)

Substitute x into equation (6)

6.41601 = 0.225 (5.34 - 2.2y)² + 0.495 y²     [expand bracket]

6.41601 = 0.225 [28.5156 - 23.496y + 4.84y²] + 0.495 y²   [remove bracket]

6.41601 = 6.41601 - 5.2866y + 1.089y² + 0.495 y²

1.584y² - 5.2866y = 0

y(1.584y - 5.2866) = 0

y = 0       or       1.584y - 5.2866 = 0

y = 0       or       1.584y = 5.2866

y = 0       or       y = 3.3375

Since y = v₂ cannot be zero because the puck will definitely move after collision, the second value of y = 3.3375 is considered.

Substitute this value into equation (m)

x = 5.34 - 2.2y

x = 5.34 - 2.2(3.3375)

x = 5.34 - 7.3425

x = -2.0025

Therefore,

x = v₁ = -2.0025m/s

y = v₂ = 3.3375m/s

(1) From the analyses above, the speed of the 0.450kg puck after collision is -2.0025m/s.

(2) Since the speed is negative, it shows that the 0.45kg puck moves opposite the direction at which it was moving before collision. It moves west.

(3) The speed of the 0.990kg puck after collision is 3.3375m/s.

(4) Since the speed is positive, it shows that the 0.990kg puck moves east. Remember that east has been taking as the positive direction.

The Type K thermocouple has a sensitivity of about 41 micro-Volts/℃, i.e. for each degree difference in the junction temperature, the output changes by 41 micro-Volts. If you have a 32-bit ADC, what is the smallest temperature change you can detect if the ADC range is 10 V?

Answers

Answer:

5.68*10^-5 °C

Explanation:

ADC is an acronym that means analog to digital conversion.

The numbers of bit in any ADC is the level of Voltage it possess.

The question states 32 bits, this means that it's total voltage levels is 2^32

Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:

Δv = 10 / (2^32)

Δv = 10 / 4.29^9

Δv = 2.33*10^-9

The thermocouple sensitivity is 41 micro volts/°C

Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change

ΔT = Δv/thermocouple sensitivity

ΔT = 2.33*10^-9 / 41*10^-6

ΔT = 5.68*10^-5 °C

Thus, the smallest temperature change is 5.68*10^-5 °C

A mass of 2.5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate the work done by the steam during this process. Use steam tables.

Answers

Answer:

the work done by the steam during this process = 119.575 kJ

Explanation:

From the information given:

At state 1

Let obtain the specific volume of the saturated vapor from the Saturated water- Pressure table A-5 at the pressure of 100 kPa

[tex]v_1 = v_{g \ 100 \ kPa)[/tex]

[tex]v_1 = 1.6941 \ m^3 /kg[/tex]

where [tex]v_1[/tex] is the specific volume of the saturated vapor at state 1.

At state 2:

From the tables A-6 of Superheated water at the pressure of 100 kPa or 0.1 MPa and at the temperature of 200°C, the specific volume [tex]v_2 = 2.1724 \ m^3 /kg[/tex]

where [tex]v_2[/tex] is the specific volume of the superheated water at state 2.

The workdone by the steam during the process can be expressed by the formula:

[tex]W = P(V_2 -V_1)[/tex]

[tex]W = mP(v_2-v_1)[/tex]

where;

m = mass of the saturated water vapor

P = pressure of the saturated water vapor

[tex]V_2 =[/tex] volume of the superheated water at state 2

[tex]V_1 =[/tex] volume of the saturated water at state 1

Replacing our values ;

W = 2.5 (100) ( 2.1724 -1.6941)

W = 250(0.4783 )

W = 119.575 kPa.m³ [tex]\times \dfrac{1 \ kJ}{1 \ kPa.m^3}[/tex]

W = 119.575 kJ

the work done by the steam during this process = 119.575 kJ

what is an atomic nucleus​

Answers

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Answers

Answer:

a

  [tex]F = 2.32*10^{-17} \ N[/tex]

b

[tex]n =3869 \ electrons[/tex]

Explanation:

From the question we are told that

   The  charge on each water drop is  [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]

   The  distance of separation is  [tex]d = 1.22\ cm = 0.0122 \ m[/tex]  

   

Generally the electrostatic force between the water drops is mathematically represented as

      [tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]

Here  k is the coulombs constant with value  [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

So  

      [tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]

       [tex]F = 2.32*10^{-17} \ N[/tex]

Generally the quantity of charge is mathematically represented as

        [tex]q = n * e[/tex]

Here n is the number of electron present

and  e is the charge on one electron with value  [tex]e = 1.60*10^{-19} \ C[/tex]

    So

         [tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]

         [tex]n =3869 \ electrons[/tex]

   

Which question cannot be answered through making measurements?
A. Should wolves be reintroduced into national parks?
B. What is the birthrate for wolves in their natural habitats?
O c. How would reintroducing wolves into a national park affect the
deer population there?
D. Which species in a national park do wolves hunt?

Answers

a. is the answer for this question

"Should wolves be reintroduced into national parks" is a question that cannot be answered through making measurements, therefore the correct answer is option C.

What is a unit of measurement?

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

A length, for instance, is a physical quantity. A defined, predetermined length is represented by the length unit known as the meter.

A qualitative type of question cannot be answered while making a quantitative measurement.

Thus, the question of whether wolves should be reintroduced into national parks cannot be answered by taking measurements, so option C is the appropriate response.

Learn more about the unit of measurement from here

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A resistor is connected in series with a power supply of 20.00 V. The current measure is 0.50 A. What is the resistance of the resistor

Answers

Answer:

40 Ω

Explanation:

The following data were obtained from the question:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

From ohm's law:

V = IR

Where:

V is the potential difference.

I is the current.

R is the resistance.

With the above formula, we can obtain the resistance of the resistor as follow:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

V = IR

20 = 0.5 × R

Divide both side by 0.5

R = 20/0.5

R = 40 Ω

Therefore, the resistance of the resistor is 40 Ω

x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering

Answers

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

[tex]2d\sin\theta=n\lambda[/tex]

For second order, n = 2

[tex]\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)[/tex]

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

Sapphire has an index of refraction of 1.77. The critical angle for sapphire in air with an index of refraction of 1.0 is _____ degrees.

Answers

Answer:

The critical angle is 34 degree.

Explanation:

Given the index refraction = 1.77

We have to find the critical angle of sapphire when the air has the index of refraction of 1.0.

Use the below formula to find the angle.

sin(θ)c = 1 / n

sin(θ) c  = 1 / 1.77

(θ) c  = Sin^-1 (0.565)

= 34.4 degree

= 34 degrees.

Thus, the critical angle is 34 degree.


If you are driving at a constant velocity of 5 m/s what is your
acceleration?

Answers

"Acceleration" means the rate at which velocity is changing.

If velocity is constant, then it isn't changing.

If velocity isn't changing, then acceleration is zero.

Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determine your own weight in newtons.

Answers

Answer:

Weight of the woman in Newton

   [tex]x = 698.6 \ N[/tex]

 Mass of the woman in slug

  [tex]Mass = 4.86 \ slug[/tex]

 Mass of the woman in kg

   [tex]Mass = 16 \ kg[/tex]

  My weight in Newton

  [tex]W = 784 \ N[/tex]

Explanation:

From the question we are told that

   The weight of the woman in pounds is  [tex]W = 157 \ lb[/tex]

Converting to Newton

    1 N  =  0.22472 lb

    x N  =  157

=>   [tex]x = \frac{157 * 1}{0.22472}[/tex]

=>   [tex]x = 698.6 \ N[/tex]

Obtaining the mass in slug

   [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 32.2 ft/s^2[/tex]

So

     [tex]Mass = \frac{157 }{32.2}[/tex]

       [tex]Mass = 4.86 \ slug[/tex]

Obtaining the mass in kilogram

     [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 9.8 \ m/s[/tex]

So

   [tex]Mass = \frac{157 }{9.8}[/tex]

   [tex]Mass = 16 \ kg[/tex]

Generally weight is mathematically represented as

     [tex]W = m * g[/tex]

Given that my mass is   80 kg   then my weight is  

     [tex]W = 80 *9.8[/tex]

      [tex]W = 784 \ N[/tex]

26500 in scientific notation

Answers

Answer:

in scientific notation 26500 is 2.65×10⁴

Answer:

2.56*10^4

26500

(We have to move the decimal 4 places)

2.6500*10^4

So the answer is 2.6500*10^4 or 2.56*10^4

Hope this helps! :)

Your friend asks you for a glass of water and you bring her 5 millilitersof water. Is this more or less than what she was probably expecting?Explain your reasoning

Answers

This is less than she was expecting because 5 milliliter of water is not a lot and she was expecting a glass of water rather than a splash of water.

With the sun and the earth back in their regular positions,consider a space probe with mass mp = 125 kg launched from the earth towardthe sun. When the probe is exactly halfway between the earth andthe sun along the line connecting them, what is the direction ofthe net gravitational force acting on the probe? Ignore the effects of other massive objectsin the solar system, such as the moon and other planets.
A. The force is toward the sun.
B. The force is toward the earth.
C. There is no net force because neither the sunnor the earth attracts the probe gravitationally at themidpoint.
D. There is no net force because the gravitationalattractions on the probe due to the sun and the earth are equal insize but point in opposite directions, so they cancel each otherout.

Answers

Answer:

A

Explanation:

Gravitational force given by

F_ps =G(m_pM_s)/r^2

G= gravitational constant

m_p=mass of the probe =125 Kg

M_s= mass of sun

r= distance between probe and Sun.

Similarly, F_pe = G(m_pM_e)/r^2.

m_p = mass of probe

M_e = mass of earth

Clearly, M_s> M_e . Therefore, F_ps> F_pe.

Hence,  A. The force is toward the sun.

A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?

Answers

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

The speed of the ball after 4s is 10 m/s.

There's not enough information to find its velocity. (We would need to know what direction it's moving.)

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?

Answers

Answer:

m = 312.5

Explanation:

Given that,

The focal length of the objecive lens, [tex]f_o=1\ cm[/tex]

The focal length of eye piece, [tex]f_e=2\ cm[/tex]

length of the tube, L = 25 cm

We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :

[tex]m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}[/tex]

D = 25 cm

So,

[tex]m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5[/tex]

So, the overall magnification of the microscope is 312.5.

Pls help answer 7 and 8

Answers

Answer:

science is the knowledge gained through observations and experimentations

Characteristics

consistency

observability

predictability

testability

tentativeness

components of nature are soil,atmosphere and radiation

Which is not an example of an external force acting on an object? (1 point)
• a meteor traveling unhindered through space
O a hockey puck striking the back of the net and coming to rest
O a billiard ball bouncing off the rail of a pool table
O a batter hitting a baseball

Answers

Answer:

A. a meteor traveling unhindered through space

Explanation:

A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum

Answers

Answer:

The change in momentum is Δp= 0.028 kg m/s

Explanation:

An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.

The change in moment is given by the expression below

Δp=m⋅(Δv)=m⋅(vf−vi) .

Given data

mass m= 0.140-kg

initial velocity vi= 120 m/s

final velocity vf= 1 m/s

substituting we have

Δp=m⋅(Δv)=0.14⋅(1−1.2)

Δp=m⋅(Δv)=0.14⋅(-0.2)

Δp= 0.028 kg m/s

The change in momentum was found to be Δp= 0.028 kg m/s

Danielle exerts a 14.0 N force to compress a spring by a distance of 8.00 cm. What is the spring constant of this spring

Answers

Answer:

175 N/m

Explanation:

Given:

Force = F=  14.0 N

Distance = x = 8.00 cm = 0.08 m

To find:

spring constant

Solution:

spring constant is calculated by using Hooke's law:

k =  F/x

Putting the values in above formula:

k = 14.0 / 0.08

k = 175 N/m

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Answers

Answer:

The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

[tex]E_{i}=E_{f}[/tex]

[tex]\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2[/tex]

[tex]\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})[/tex]

[tex]0.00075(400-v_{f}^2)=0.18 [/tex]

[tex]400-v_{f}^2=\dfrac{0.18}{0.00075}[/tex]

[tex]-v_{f}^2=240-400[/tex]

[tex]v_{f}^2=160[/tex]

[tex]v_{f}=4\sqrt{10}\ m/s[/tex]

Hence, The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the +x -direction with magnitude 3 m/s^2 < a < 6  m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?

Answers

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

We want to find the range of possible positions for the cart at t = 1s.

The range is:

14.5m ≤ p ≤ 16m

Finding the cart's motion equations.

First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:

a(t) = a

To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.

v(t) = a*t + 3m/s

To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.

p(t) = (a/2)*t^2 + (3m/s)*t + 10m

Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.

The minimum acceleration is a = 3m/s^2, so we have the minimum position:

p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m

The maximum acceleration is 6m/s^2, so the maximum position is:

P(1s) =  (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m

So the range of the position at t = 1s is:

14.5m ≤ p ≤ 16m

If you want to learn more about motion equations, you can read:

https://brainly.com/question/2473092

A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator(a) is standing still.(b) is moving upward at constant velocity.(c) is accelerating upward.(d) is moving downward at constant velocity.(e) is accelerating downward.

Answers

Answer:

(c) is accelerating upward

Explanation:

When a person stand in an elevator moving upwards, he feels heavier because the elevator's floor presses harder on his feet, and the scale will show a higher reading than when the elevator is at rest.

According to Newton's second law

R = mg + ma

where;

R is the reading of the scale = apparent weight of the person

mg is the normal weight of the person

ma is the upward force acting on the person

Therefore, his apparent weight will be the greatest when the elevator is accelerating upward

(c) is accelerating upward

What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Answers

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m

Hence

2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m

A wire is 24.5 m long at 4.00°C and is 1.40 cm longer at 34.0°C. Find the wire's coefficient of linear expansion (in (°C)−1)

Answers

Answer:

1.9×10⁻⁵ °C⁻¹

Explanation:

Coefficient of linear expansion: This can be defined as as increase in length of a material, per unit length per degree rise in temperature, The S.I unit is (K⁻¹).

From the question above,

α = ΔL/L(T₂-T₁)................... Equation 1

Where α = coefficient of linear expansion, ΔL = Change in length, L = original length, T₁ = Initial Temperature, T₂ = Final temperature.

Given: ΔL = 1.4 cm = 0.014 m, L = 24.5 m, T₁ = 4.00°C, T₂ = 34°C

Substitute these values into equation 1

α = 0.014/[24.5(34-4)]

α = 0.014/735

α = 1.9×10⁻⁵ °C⁻¹

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