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4 years ago

1. Which mathematical representation correctly identifies impulse?

Physics

1 answer:

horsena [70]4 years ago

3 0

Answer:

1. B. Impulse = Force × Time

2. A. The momentum of each ball changes, and the total momentum stays the same

3. -55 kg·m/s

4. B. 3.5 kg

5. C. 6.3 m/s

Explanation:

1. The impulse is the momentum change of an object due to a force applied for a given period

2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;

The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same

3. By the principle of conservation of linear momentum, we have;

The sum of the momentum before the collision = The sum of the momentum after collision

Given that the objects move together after the collision, the total momentum is therefore;

Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s = -55kg·m/s

4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s

Where;

m₁ = The mass of the first object

m₂ = The mass of the second object

The total momentum of the system = 250 kg·m/s

From momentum = Mass × Velocity, we have;

Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg

The mass (m₁ + m₂) = 5.0 kg

Given that m₁ = 1.5 kg, we have;

m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg

The mass of the second object = 3.5 kg

5. The mass of the cue stick = 0.5 kg

The velocity of the cue stick = 2.5 m/s

The mass of the ball = 0.2 kg

The initial velocity of the ball = 0 m/s

Given that total initial momentum = Total final momentum, we have;

0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0

0.5 kg × 2.5 m/s = 0.2 kg × v

v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25 m/s ≈ 6.3 m/s

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At what distance along the z-axis is the electric field strength a maximum?

Lesechka [4]

The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is

dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) 

where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. 

However, cos(T) = x/sqrt(x^2 + R^2) 

so the equation becomes 

dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) 

dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 

Integrating around the ring you get 

E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 

E = (R/2*e0)*x*(x^2 + R^2)^-1.5 

we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is 

dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} 

dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} 

to find the maxima set this = 0, giving 

(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 

mult both side by (x^2 + R^2)^2.5 to get 

(x^2 + R^2) - 3*x^2 = 0 

-2*x^2 + R^2 = 0 

-2*x^2 = -R^2 

x = (+/-)R/sqrt(2)

4 0

4 years ago

One-half mole of propane combines with 2.5 moles of oxygen to produce_____moles of water and_____moles of carbon dioxide.

Virty [35]

The balanced equation of the combustion of the propane is

$C_3H_8+5O_2\rightarrow3CO_2+4H_2O$

From the above chemical equation, it is clear that 1 mole of propane requires 5 moles of oxygen to undergo complete combustion as the result of this reaction, 3 moles of carbon-di-oxide and 4 moles of water molecules will be produced.

Therefore one-half mole of propane requires 2.5 moles of oxygen to undergo complete combustion. And in the reaction, 1.5 moles of carbon-di-oxide and 2 moles of water molecules are produced.

Therefore the answer is, 'One-half of propane combines with 2.5 moles of oxygen to produce 2 moles of water and 1.5 moles of carbon dioxide'

7 0

1 year ago

The top speed you will ever need to go in a parking lot is_

Kruka [31]

10 mph i jus got it right on the test

3 0

3 years ago

A student said, "The displacement between my dorm and the lecture hall is 1 kilometer."

Ann [662]

Answer:

d) Both a and b are correct

Explanation:

Displacement: It is defined as the distance between initial and final position during motion.

Distance: It is defined as the total path length traveled by object

It is the distance of one place from other place.

Student said that the displacement between my dorm and the lecture hall is 1km.

It is not displacement .It is distance or we say path length.

Therefore, he is using incorrect physical quantity for the information provided.

He should have called distance 1 km or path length 1 km.

Option d is true.

5 0

3 years ago

Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha

Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y= F₂ cos 32° j

F₂y= 35 x cos 32° j = 29.68 j

The total force F

F= 33.47 i + 18.54 i - 55.71 j + 29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

$F=\sqrt{52.01^2+26.03 ^2}\ N$

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0

4 years ago