Very large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge the likelihood of injury is the severity index ( ), defined as =5/2 . In the expression, is the duration of the accleration, but is not equal to the acceleration. Rather, is a dimensionless constant that equals the number of multiples of that the acceleration is equal to.

In one set of studies of rear-end collisions, a person's velocity increases by 12.1 km/h with an acceleration of 35.0 m/s2 . Let the + direction point in the direction the car is traveling. What is the severity index for the collision?

How far does the person travel during the collision if the car was initially moving forward at 5.80 km/h ?

Answer:

increasing the separation between the plates

Explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV  V = VOLTAGE , c = capacitance  

E = 1/2 eAV^2/ D  ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

Answer:

Explanation:

The box is at rest because , the component of weight along the surface downwards is equal to frictional force .

frictional force = mgsinθ

= 20 x 9.8 sin 30

= 98 N .

Answer:

He needs clay gravel and rocks with cracks

Answer:

4.The atoms are vibrating in place

Explanation:

The answer is;  The atoms are vibrating in place

We know that, molecules in the crystal have a definite position in the crystal and are bonded to each other by electrostatic forces. However, since the molecules have some energy, they vibrate in their positions. Their energy, however, is not high enough to cause them to overcome the strong bonding (unless the crystal is heated or the atoms are irradiated).

Answer:

L= 276.4 mm

Explanation:

Given that

E= 180 GPa

d= 3.7 mm

F= 1890 N

ΔL= 0.45 mm

We know that ,elongation due to load F in a cylindrical bar is given as follows

[tex]\Delta L =\dfrac{FL}{AE}[/tex]

[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]

Now by putting the values in the above equation we get

[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]

L=0.2764 m

L= 276.4 mm

Therefore the length of the specimen will be 276.4 mm

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 = 2.5 m / s²

Explanation:

This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction

Let's write the second law for bodies in the inclined plane

    W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a

    N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0

    N₁ + N₃ = W_{1y} + W_{3y}

let's use trigonometry to find the weight components

    sin 30 = Wₓ/ W

    Wₓ = W sin 30

    cos 30 = W_{y} / W

    W_{y} = W cos 30

we substitute

    N₁+ N₃ = W₁ cos 30 + W₃ cos 30

    W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a

     a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)

     a = g sin 30 - μ g cos30

let's calculate

     a = 9.8 sin 30 - 0.28 9.8 cos 30

     a = 4.9 - 2,376

     a = 2.5 m / s²

Answer:

  48.54 m/s

Explanation:

If the rock takes 9 seconds to reach your position after being thrown, it reaches its maximum height in 4.5 seconds.

The height the rock reaches above your position is ...

  h = (1/2)gt^2 = (4.9 m/s^2)(4.5 s)^2 = 99.225 m

This height is an additional 21 m above the water, so the maximum height above the water is ...

  99.225 m +21.0 m = 120.225 m

The velocity (v) achieved when falling from this distance is found from ...

  v^2 = 2gh

  v = √(2(9.8)(120.225)) = √2356.41 ≈ 48.543 . . . . m/s

The speed of the rock when it hits the water is about 48.54 m/s.

Answer:

0.25916 lb-s/ft^2

Explanation:

Given:-

- The specific gravity of oil, SGo = 0.8

- The specific gravity of sphere, SGo = 8

- Terminal velocity of sphere, v = 2.5 fpm

- The diameter of sphere, D = 0.25 in

Find:-

What is the viscosity of the oil?

Solution:-

- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).

- Develop a free body diagram for the sphere. There are forces acting on the sphere.

- The downward acting force is due to the weight of the sphere ( W ):

                         [tex]W = m_s*g[/tex]

Where,

The mass ( m_s ) of the sphere is given as:

                          [tex]m_s = S.G_s*p_w*V_s[/tex]

Where,

              ρ_w : Density of water = 1.940 slugs/ft3

               V_s: The volume of object ( sphere )

- The volume of sphere is expressed as a function of radius:

                        [tex]V_s = \frac{\pi *D^3}{6}[/tex]

Hence,

                        [tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]

- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.

- The buoyant force ( Fb ) is given by:

                    [tex]F_b = S.G_o*p_w*V_s*g[/tex]

- Therefore the buoyant force ( Fb ) becomes:

                    [tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]

- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.

- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:

                    [tex]F_d = 3*\pi*D*u*v[/tex]

Where,

                    μ: The viscosity of fluid

                    v : The velocity of object

Therefore,

                       [tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]

- Apply Newton's second law of motion for the sphere travelling in the fluid:

                      [tex]F_n_e_t = m_s*a[/tex]

Where,

                     a: Acceleration of object = 0 ( Terminal velocity condition )

                     [tex]F_n_e_t = 0[/tex]

- Plug in the three forces acting on the metal sphere:

                     [tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]

Answer:

Sorry but i dont know

Explanation:

Answer:

tug_tug = 570 10³ l

Explanation:

In this problem, each propeller creates a force that makes the boat rotate, so the tugs have to create a die of equal magnitude rep from the opposite direction

           ∑ τ = 0

           F1 la+ (-F1) (-l) = τ-tug

           τ-tug = 2 f1 l

            τ-tug = 2 28510³ l

            tug_tug = 570 10³ l

where the is the distance from the propane axis to the point where the ship turns

This force may be less depending on where the tug is.

B. Formation of new stars

Answer:

Explanation:

The problem is based on Newton's law of cooling .

According to Newton's law

dQ / dt = k ( T - T₀ ) ,

dT / dt = k' ( T - T₀ )          ; dT / dt is rate of fall of temperature.

T is average  temperature of hot body , T₀ is temperature of surrounding .

In the first case rate of fall of temperature = (210 - 191) / 2.5

= 7.6 degree / s

average temperature T = (210 + 191) /2

= 200.5  

Putting in the equation

7.6 = k' ( 200.5  - 64 )

k' = 7.6 / 136.5

= .055677

In the second case :---

In the second case, rate of fall of temperature = (191 - 156) / t  

= 35 / t   , t is time required.

average temperature T = (156 + 191) /2

= 173.5  

Putting in the equation

35 / t = .05567 ( 173.5 - 64 )

t = 5.74 minute .

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]

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

[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]

[tex]7.2 =200*10^{-3}*v^{2}[/tex]

[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]

[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

Answer:

b

Explanation:

Answer:

The correct answer is D, diatoms have glass-like cell walls.

Answer:

Explanation:

mass of probe m = 474 Kg

initial speed u = 275 m /s

force acting on it F = 5.6 x 10⁻² N

displacement s = 2.42 x 10⁹ m

A )

initial kinetic energy = 1/2 m u²  , m is mass of probe.

= .5 x 474 x 275²

= 17923125 J  

B )

work done by engine

= force x displacement

= 5.6 x 10⁻² x 2.42 x 10⁹

= 13.55 x 10⁷ J  

C ) Final kinetic energy

= Initial K E + work done by force on it

= 17923125 +13.55 x 10⁷

= 1.79 x 10⁷ + 13.55 x 10⁷

= 15.34 x 10⁷ J

D ) If v be its velocity

1/2 m v² = 15.34 x 10⁷

1/2 x 474 x v² =  15.34 x 10⁷

v² = 64.72 x 10⁴

v = 8.04 x 10² m /s

= 804 m /s

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]

the relative velocity is:

[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]

Next, you use the general for of a wave:

[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]

you take the amplitude as 2.0/2 = 1.0m.

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]

Answer:

a) Q = 1436 kW

b) P ≈ 776 kW

Explanation:

Let's begin by listing out the given parameters:

T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,

V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s

To solve the question, we make this assumption that the size of the pipe is constant

a) No change in velocity implies that heat added is isochoric

Q = m * C * ΔT

Cv of air at 300 K(≈20 °C) = 0.718

Q = 2.5 * 0.718 * (820 − 20)

Q = 1436 kW

b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql

V2² - V3² = 55² - 40² = 1425

ΔT = T2 - T3 = 820 - 510 = 310 °C

Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK

Ql = 4.3 kW

P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3

P = 778.875 + 1.78125 - 4.3 = 776.35625

P ≈ 776 kW

Answer:

a and b are the correct answers

Explanation:

Answer:

A)  physical property; mass.

Explanation:

took the test

Answer: (a) and (b) => check attached file.

(c). Picture (a) and (b) will both remain the same.

Explanation:

IMPORTANT: The solution to the question (a) and (b) that is  (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.

It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.

Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.

Hence, the solution to option C is given below;

(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.

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: R.A.M

Explanation:

Answer:

Explanation:

momentum of sedan of 1600 kg = 1600x v , where v is its velocity

momentum of suv of 2300 kg = 2300 x u where u is its velocity .

force of friction = ( 1600 + 2300 ) x 9.8 x  .75 ( fiction = μ mg )

= 28665 N

distance by which friction acted = √ (5.54² + 6.19²)

= 8.3 m

work done by friction

= 28665 x 8.3

= 237919.5 J

Total kinetic energy of cars = work done by friction

1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5

16 v² + 23 u² = 4758.4

1600 x v / 2300 u = 6.19 / 5.54

v / u = 1.6

v = 1.6 u

putting this equation in fist equation

40.96 u² + 23 u² = 4758.4

= 63.96 u² = 4758.4

u² = 74.4

u = 8.62 m /s

v = 13.8  m /s

Answer:

When a current carrying wire is in your left hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current

Explanation:

This is in line with the left hand rule

B.) When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.

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:

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:

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:

Considering that there is no obstructions, he could go west from the start.

Explanation: