Which of the following is probably not a reason why society might initially reject a new scientific theory

The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

tatuchka [14]

By Newton's second law, assuming F is horizontal,

• the net horizontal force on the larger block is

F - µmg = 3mA

where µmg is the magnitude of friction felt by the larger block due to rubbing with the smaller one, µ is the coefficient of static friction between the two blocks, and A is the block's acceleration;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3mg is the weight of the larger block, and mg is the weight of the smaller block;

• the net horizontal force on the smaller block is

µmg = ma

where µmg is again the friction between the two blocks, but notice that this points in the same direction as F. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net vertical force on the smaller block is

mg - mg = 0

where mg is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for A and a :

A = (F - µmg) / (3m)

a = µg

5 0

1 year ago

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

solution:

given:

p_o=1.07*10^5 N/m^2

ρ_infinity=1.23 kg/m^2

p_infinity=1.01*10^5 N/m^2

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

2 years ago