A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Answer:

a) 8.67m

b) 1000N

Explanation:

(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:

[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]

once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:

[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]

v: final velocity=0

vo: initial velocity = 72km/h = 60 m/s

by replacing the values of these parameters you obtain for x:

[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]

(b) The horizontal component of the force exerted by the trailer hitch is given by:

[tex]F_T=5000N-4000N=1000N[/tex]

Answer:1. The masses Of The objects 2.the distance between the objects

Explanation:

newton's law of universal gravitation states that the force of attraction between two masses is directly proportional to the product Of The masses, and inversely proportional to thermal square Of The distance between them.therefore we can conclude that the two factors affecting the force between masses are:(1)the masses of the objects (2)the distance between the objects

The two factors that affect the force between two masses, according to the universal law of gravitation are

Newton's Universal law of gravitation states that the force of attraction that exists between particle or objects is directly proportion to the product of their masses and inversely proportional to the distances between them .

Therefore, The two factors that affect the force between two masses, according to the universal law of gravitation are

Learn more about Newton's universal law of gravitation below.

https://brainly.com/question/10454047

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Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

Answer:

The answer is A

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent

Answer:

A.

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent.

Answer:

D. the linear velocity of the point of contact (relative to the inclined surface) is zero

Explanation:

The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due  point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction  becomes zero . Hence work done by friction becomes zero.

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

[tex]a=cs[/tex]   (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

[tex]v^2=v_o^2+2a \Delta s[/tex]   (2)

vo: initial  velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

[tex]v^2=2(cs)\Delta s[/tex]

You do the constant c in the last equation, then you replace the values of v, s and Δs:

[tex]c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44[/tex]

Answer:

[tex]W=5.16 J[/tex]  

Explanation:

Using the Hooke's law we can find the elasticity constant:

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

[tex]30.8=-k*0.177[/tex]

[tex]k=|-\frac{30.8}{0.177}|[/tex]

[tex]k=174 N/m[/tex]

Now, we know that the work done is equal to the elastic energy, so we will have:

[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]

x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)

x1 is the initial distance (x1 = 0.177 m)

[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]

[tex]W=5.16 J[/tex]    

I hope it helps you!

Answer: what ( younger brothers of hers )

Explanation:

K bye

Answer:

Explanation:

moment of inertia of each blade which is similar to rod rotating about its one end

= 1/3 ml²

moment of inertia of 3 blades = ml²

= 5500 x 46²

I = 11638 x 10³ kg m²

angular velocity = 2πn where n is rotation per second

n = 11 / 60

angular velocity = 2π x 11/60

= 1.1513 rad /s

angular momentum

= moment of inertia x angular velocity

=  11638 x 10³ x 1.1513

= 13399 x 10³ kg m² per second.

Explanation:

P=E/T

E=15N

T=2s

P=15/2

P=7.5

Answer:

During an earthquake, seismic waves travels through the Earth's interior as body or p waves.

Explanation:

If neither of the bold words look familiar from your lesson feel free to ignore this answer

Answer:

Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42

The image is a virtual and upright image.

Explanation:

The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.

The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.

Given f = -33.0cm, v = -19.0cm

using thr mirror formula to get the object distance u, we have;

[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]

To calculate the image height, we will use the magnification formula

M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]

M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]

Given Hi = 7.0cm

v = 19.0cm

u = 44.8cm

HI = 7*44.8/19

HI = 16.5cm

The object height is 16.5cm

Magnification = v/u = 19.0/44.8 = 0.42

SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image

Answer:

A black hole

Explanation:

I'm not sure, i just know that's the right answer

Answer:

coefficient of friction (μ) and normal force (N)

Answer: How hard the surfaces push together and the types of surfaces involved

Explanation:

Answer:

Magnetic field can be used to produce current, infact a changing magnetic field can produce current.

A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.

For a loop of area A and resistance R.

I =dPhi/dt/R

В. А

I = AcosФ/R .dB /dt

But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).

We were unable to transcribe this image

Ф= В. А

I = Acos dB Rd

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

Explanation:

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex]   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

[tex]\int EdS=ES=E(4\pi r^2)[/tex]   (2)

Qin is calculate by using the charge density:

[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex]  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]

Next, you use the results of (3), (2) and (1):

[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]

hence, the electric field is 1580594.95 N/C

Answer:

There will be no change in the angular momentum of the system.

Explanation:

Total angular momentum of the system  will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .  

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]

Force applied to stop the car = 1,250 N

Answer:

a magnified and erect virtual is found to be formed.

Answer:

The total resistance is more than the largest resistor.

Explanation:

In series connection of resistors, the total resistance of the connection is the sum of all the resistance of the resistors connected.

For example; let assume three resistors with resistance 3,4 and 5 ohms, the total resistance is;

Rt = 3+4+5 = 12 ohms

So, the total resistance is greater than the largest resistor since 12 ohms is greater than 5 ohms(largest resistor).

Answer:

.7934[tex]m/s^{2}[/tex]

Explanation:

Acceleration = change in velocity / change in time

A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]

A = .7934[tex]m/s^{2}[/tex]

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

Answer:10 ohms

Explanation:

Current=0.15A

Voltage=1.5v

Resistance=voltage ➗ current

Resistance=1.5 ➗ 0.15

Resistance=10 ohms

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Answer:

Q' = 3.21*10^{-5}C

Explanation:

To find the new magnitude of the charge you take into account  that the voltage of the this capacitor is given by:

[tex]\frac{Q}{\epsilon_o\epsilon_r A}=\frac{V}{d}\\\\V=\frac{Qd}{\epsilon_o\epsilon_r A}[/tex]

Q: total charge

d: distance between parallel plates

A: area of the plates

εr: dielectric constant

εo = dielectric permittivity of vacuum

for the case of the air εr = 1, then,

[tex]V=\frac{Qd}{\epsilon_o A}[/tex]     (1)

When a dielectric material is placed in between the plates, you have, for the same voltage, and for a different charge:

[tex]V=\frac{Q'd}{\epsilon_o\epsilon_rA}[/tex]   (2)

you equal the equation (1) and (2) and obtain:

[tex]\frac{Qd}{\epsilon_o A}=\frac{Q'd}{\epsilon_o \epsilon_r A}\\\\Q'=\epsilon_r Q[/tex]

by replacing you obtain:

[tex]Q'=(7.74)(4.15*10^{-6}C)=3.21*10^{-5}C[/tex]

Answer:

a) [tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]

b) [tex]p=2.81*10^{-16} kg*m/s[/tex]    

Explanation:

a) When we have two particles traveling in parallel directions, the formula for relative velocity is:

[tex]\vec{v}_{12}=\frac{\vec{v}_{1}-\vec{v}_{2}}{1-\frac{\vec{v}_{1}\vec{v}_{2}}{c^{2}}}[/tex]

Here we have that v(1) = -v(2), the speed of the of the second particle is the negative of the first one.

If we use these equivalence we have:

[tex]\vec{v}_{12}=\frac{2\vec{v}_{1}}{1+\frac{\vec{v}_{1}^{2}}{c^{2}}}[/tex]  

[tex]\vec{v}_{12}=\frac{2*2.19*10^{8}}{1+\frac{2.19*10^{16}}{3*10^{16}}}[/tex]  

[tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]          

And, [tex]\vec{v}_{21}=-2.86*10^{8} m/s[/tex]  

b) The relativistic momentum equation to one particle observed by the other particle, is:

[tex]p=\gamma mv[/tex]

Where gamma is:

[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

Then gamma will be:

[tex]\gamma=\frac{1}{\sqrt{1-\frac{(2.86*10^{8})^{2}}{(3*10^{8})^{2}}}}[/tex]

[tex]\gamma=3.31[/tex]

Finally, the value of the momentum will be:

[tex]p=3.31*2.97*10^{-25}*2.86*10^{8}[/tex]    

[tex]p=2.81*10^{-16} kg*m/s[/tex]    

I hope it helps you!                        

Answer:

a and b are the correct answers

Explanation:

Answer:

A)  physical property; mass.

Explanation:

took the test

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

Answer:

the answer is a

Explanation:

Answer:

A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.

Explanation:

Answer:

C

Explanation:

:)))