In the equation Δp=m(vf−vi), which quantity is considered to be constant?

What is the mass of a large dog that weighs 441 newtons

PtichkaEL [24]

Weight, w = mg. g ≈ 9.8 m/s². m = mass in kg. w is weight in N

441 N = m* 9.8

9.8m = 441

m = 441/9.8

m = 45 kg.

Mass of the dog is = 45 kg

5 0

2 years ago

Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n

Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

FOR FRIEND ON MOON:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

FOR ME ON EARTH:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

Therefore, the friend on moon will be richer.

7 0

2 years ago

The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0

2 years ago

In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad

Crank

Answer:

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

Q = 1.57 10⁻² √(12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √(2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s

5 0

2 years ago

Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c

ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0

2 years ago