A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater.a) Use work and energy to find the skater's speed after gliding 100 m in this wind.b) What is the minimum value of Ug that allows her to continue moving straight north?

Answers

Answer 1

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

[tex]W = F\times d[/tex]

[tex]KE = 0.5\times m\times v^2[/tex]

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

[tex]W = F \times d = 2.62 N \times 100 m[/tex]

[tex]W = 261.6 N\times m[/tex]

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]

Now solve for v2

[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]

[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

[tex]Ff = Us\times N[/tex]

Now solve for Us

[tex]= \frac{Ff}{N}[/tex]

[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]

= 0.00494


Related Questions

¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?

Answers

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]

distancia= 25 metros

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo

Usando la tasa de conversión;

1000 m = 1 km

1 hora = 3600 segundos

[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]

Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

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I really need help with this question someone plz help !

Answers

Answer:D

Explanation:

Given

Same force is applied to each ball such that all have different masses

and Force is given by the product of mass and acceleration

[tex]F=m\times a[/tex]

[tex]a=\frac{F}{m}[/tex]

So acceleration of ball A

[tex]a_A=\frac{F}{0.5}=2F[/tex]

acceleration of ball B

[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]

acceleration of ball C

[tex]a_C=\frac{F}{1}=F[/tex]

acceleration of ball D

[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]

It is clear that acceleration of ball D is least.

A ship can float on water as long as it weighs less than water.
O A. True
O B. False

Answers

Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

What is a substance?

Answers

a particular kind of matter with uniform properties.

The self-referencing effect refers to ________.

Answers

The self-reference effect is the tendency an individual to have better memory for information that relates to oneself than information that is not personally relevant.

Which of the following is analogous to the pipes in an electrical circuit?

A. capacitors storing the incoming charge from the battery
B. large resistors causing restrictions to the flow of charge
C. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery electric current being forced uphill by the battery
D. electric current being forced uphill by the battery back to the positive terminal

Answers

The correct answer is D. electric current being forced uphill by the battery back to the positive terminal.

What is Electric Current?

Electric current is the flow of electric charge through a conducting medium, such as a wire, due to the movement of electrons or ions. The flow of charge is typically caused by the presence of an electric field that creates a potential difference (voltage) between two points in a circuit

In an electrical circuit, pipes are analogous to wires or conductive paths that allow the flow of electric current. The flow of electric current is from the positive terminal of the battery to the negative terminal, which is opposite to the direction of conventional current flow. Therefore, option C is incorrect.

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A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Calculate the state of plane stress at point C located 50 mm below the top of the beam and 0.5 m to the right of point A. Also find the principal stresses and the maximum shear stress at C. Neglect the weight of the beam.

Answers

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]

Calculate the first moment of area below point C

[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]

Now calculate shear stress at point C

[tex]=\frac{FQ}{It}[/tex]

[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]

Calculate the principal stress at point C

[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]

Calculate the maximum shear stress at piont C

[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]

How do you convert 1.3*10^6cal into joules

Answers

Answer:

5.4×10⁶J

Explanation:

1 cal = 4.184 J

1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?

Answers

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.

Given:

diameter of the circle = 1.7 m

radius f the circle would be = 1.7/2 = 0.85 m

time taken for one revolution t = 1.2 s

This rotation exercise can be treated using the rotation kinematics.

Angular acceleration:

θ = w₀ t + ½ α t²

t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)

 =>  θ = 0 + ½ α t²

 => α = 2θ / t²

=>  α= 2 × 2π / 1.2²

 => α = 4π = 8.7266 rad / s²

Let's calculate the angular velocity:

=> w = wo + α t

=>  w = 0 + α t

=> w = 8.7266 × 1.2

=> w = 10.47192 rad / s

The relationship between linear and angular velocity is

=> r = d / 2

=> r = 1.7 / 2 = 0.85 m

=> v = w r

=> v = 10.47192 × 0.85  

=> v = 8.90 m / s

Thus, the correct speed would be - 8.90 m/s

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A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?

Answers

Answer:

0.938 seconds

Explanation:

For the ball thrown upwards, we use the formula below to solve it:

[tex]s = ut - \frac{1}{2}gt^2[/tex]

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

Let x be the height at which both balls are level, this means that:

=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)

For the ball dropped downwards, we use the formula below:

[tex]s = ut + \frac{1}{2}gt^2[/tex]

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=> [tex]18 - x = 0 + 4.9t^2[/tex]

=> [tex]x = 18 - 4.9t^2[/tex]__________(2)

Equating both (1) and (2):

[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]

They will be level after 0.938 seconds

If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.

Answers

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s

Answers

Answer:

I dont know bro

Explanation:

Ask an expert

Answer:

Time=90s

Explanation:

Speed=distance /time

[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]

To move a large crate across a rough floor, you push on it with a force at an angle of 15 degrees below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.49.

Answers

Answer:

663N

Explanation:

We need to find the force that will overcome the frictional force.

The angle of the normal force is 15°.

The mass of the crate is 32 kg

The coefficient of static friction is 0.49

Frictional force is given in terms of Normal force as:

F = μNcosθ

where μ = coefficient of static friction

N = normal force

θ = angle of normal force

Frictional force is given as:

F = mg

=>mg = μNcosθ

=> N = mg/(μcosθ)

N = (32 * 9.8) / (0.49 * cos15)

N= 313.6 / 0.473

N = 663 N

The force needed to cause the box to move must be 663N or greater.

When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation", produces a sound pulse---the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside, at a distance of 0.29 m from your ear. If the pulse has a sound level of 61 dB at your ear, what is the rate at which energy is produced by the cavitation

Answers

Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?

Explanation:

The only force acting on a 3.2 kg canister that is moving in an xy plane has a magnitude of 6.7 N. The canister initially has a velocity of 3.3 m/s in the positive x direction, and some time later has a velocity of 6.9 m/s in the positive y direction. How much work is done on the canister by the 6.7 N force during this time

Answers

Answer:

The work done by the force is  5.76 J

Explanation:

Given;

mass of canister , m = 3.2 kg

magnitude of force, f = 6.7 N

initial velocity of the canister on x-axis,  [tex]v_i[/tex]= 3.3i m/s

final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s

The work done on the canister = change in the kinetic energy of the canister

[tex]W = K.E_f - K.E_i[/tex]

where;

K.Ei is the initial kinetic energy

K.Ef is the final kinetic energy

The initial kinetic energy:

[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]

The final kinetic energy:

[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]

W = 11.04 - 5.28

W = 5.76 J

Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J

A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secondary coil is connected to a light bulb that operates on 120 V. Determine the ratio of the secondary to primary turns needed for the bells transformer. Determine the ratio of the secondary to primary turns needed for the bells transformer. Ns/Np=?

Answers

Answer:

N₂ / N₁ = 13.3

Explanation:

A transformer is a system that induces a voltage in the secondary due to the variation of voltage in the primary, the ratio of voltages is determined by the expression

           ΔV₂ = N₂ /N₁  ΔV₁

where ΔV₂ and ΔV₁ are the voltage in the secondary and primary respectively and N is the number of windings on each side.

In this case, they indicate that the primary voltage is 9.0 V and the secondary voltage is 120 V

therefore we calculate the winding ratio

         ΔV₂ /ΔV₁ = N₂ / N₁

         N₂ / N₁ = 120/9

         N₂ / N₁ = 13.3

s good clarify that in transformers the voltage must be alternating (AC)

1. Which of the following is NOT a vector quantity? (a) Displacement. (b) Energy. (c) Force. (d) Momentum. (e) Velocity.

Answers

Answer:

B. energy

Explanation:

A vector has direction.

Energy does not have a direction.

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.

Answers

Answer:

a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g,  y = 24.25 m

e) R = 138.46 m

Explanation:

This is a projectile launch exercise

a) let's use trigonometry to find the components of the initial velocity

  cos θ = v₀ₓ / v₀

  v₀ₓ = v₀ cos θ

   

v₀ₓ = 38 cos 35

v₀ₓ = 31.13 m / s

b) sin θ = [tex]v_{oy}[/tex] / v₀

    v_{oy} = v₀ sin θ

    v_{oy} = 38 sint 35

    v_{oy} = 21 80 m / s

c, d) to find the maximum height, the vertical speed is zero

     v_{y}² = v_{oy}² - 2 g y

     0 = [tex]v_{oy}[/tex]² - 2 gy

     y = v_{oy}² / 2g

let's calculate

     y = 21.80 2 / (2 9.8)

     y = 24.25 m

e) They ask to find the horizontal distance

    for this we can use the expression of reaches

       R = v₀² sin 2θ / g

let's calculate

      R = 38² sin (2 35) / 9.8

       R = 138.46 m

Which of the following is NOT true about main sequence stars?

A.) The forces of gravity and nuclear fusion balance so the star does not collapse or explode.

B.) Temperature is directly related to brightness.

C.) The forces of gravity and nuclear fusion are not in balance so the star's core collapses while the outer layers expand.

D.) Temperature is related to size.

Answers

Answer:

the statements the B is not true

Explanation:

In the stars the force of gravity tends to collapse them, by joining the atoms nuclear reactions that create an outward force until the two forces reach an equilibrium

The temperature of a star is a reflection of the energy within it and it is related to the intensity of the nuclear rations and not to the size of the stars

In examining the statements the B is not true

5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities

Answers

Insoluble heavy impurities

Answer:

D

Explanation:

Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.

Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.

With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.

Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.

What caused the disappearance of land bridges?
A. Volcanic outgassing
B. Shrinking of the polar ice caps
C. Beginning of an ice age
D. A mass extinction​

Answers

Answer: B

Explanation:

I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges

Answer:B :)

Explanation:

A 328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at 13.0 m/s in the same direction. If the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision

Answers

Answer:

14.04 m/s

Explanation:

To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

We have the following data:

m1 = m1' = 328,

m2 = m2' = 790,

v1 = 19.1,

v2 = 13,

v2' = 15.1.

Using this data, we can find v1' (final velocity of the first car):

328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1

16534.8 = 328 * v1' + 11929

328 * v1' = 4605.8

v1' = 14.04 m/s

A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answers

Answer:

Spring Constant = 279.58 N/m

Explanation:

We are given;

Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg

Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m

Maximum height; h = 4.46 m

The formula for the energy in the spring is given by;

E = ½kx²

where:

k is the spring constant

x is the distance the spring is compressed.

Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;

½kx² = mgh

Plugging in the relevant values, we have;

½ * k * 0.0801² = 0.0205 * 9.81 * 4.46

0.003208005k = 0.8969

k = 0.8969/0.003208005

k = 279.58 N/m

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 810 N and a radius of 1.56 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s

Answers

Answer:

Kinetic Energy of the disk = 252 J

Explanation:

weight of disk = 810 N

radius = 1.56 m

applied force = 49 N

time = 2.95 s

kinetic energy of disk = ?

first, we find the mass of the disk

mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2

mass of disk = 82.57 kg

torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m

moment of inertia I = m[tex]r^{2}[/tex] = 82.57 x [tex]1.56^{2}[/tex] = 200.9 kg-[tex]m^{2}[/tex]

recall that

Torque T = Iα

where α = angular acceleration

76.44 = 200.9α

α = 76.44/200.9 = 0.38 m/s^2

from the equation of angular motion,

ω = ω' + αt

where ω =  final angular speed

ω' = initial angular speed = 0 rad/s since disk starts from rest

t = time = 2.95 s

imputing values into the equation, we have

ω = 0 + (0.38 x 2.95)

ω = 1.12 rad/s

kinetic energy of the disk = I[tex]w^{2}[/tex]

KE = 200.9 x [tex]1.12^{2}[/tex]

Kinetic Energy of the disk = 252 J

What is the power of a child that has
done work of 50J in 10 seconds.
(a)50W (b)20W (c)30W (d)5W​

Answers

_____________________________

Solution,

Work=50 Joule

Time=10 seconds

Power=?

Now,

Power=Work/time

= 50/10

= 5 Watt.

So the right answer is 5 W

Hope it helps..

Good luck on your assignment

__________________________

A projectile is defined as

Answers

Answer:

By definition, a projectile has a single force that acts upon it - the force of gravity.

Explanation:

A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

// have a great day //

first law of equilibrium

Answers

Answer:

for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.

Hope I helped

Answer:

An object in static equilibrium has zero net force acting upon it.

The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as

 F = F1+ F2+ F3+ F4+. . . = 0

small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? It is impossible to tell since the velocities are not given. The truck experiences the greater average force. It is impossible to tell since the masses are not given. The small car and the truck experience the same average force. The small car experiences the greater average force.

Answers

Answer:

The correct option is D: "The small car and the truck experience the same average force."

Explanation:

The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.

Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?

Answers

Answer:

75.0 cm

Explanation:

becouse i don,t no the right answer

A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity

a. The potential at the center of the sphere is zero.
b.The potential is lowest, but not zero, at the center of the sphere.
c. The potential at the center of the sphere is the same as the potential at the surface.
d. The potential at the center is the same as the potential at infinity.
e. The potential at the surface is higher than the potential at the center.

Answers

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex]      inside the sphere

[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex]      for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

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