All categories

2 years ago

A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

Physics

1 answer:

never [62]2 years ago

3 0

Answer:

a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

F₁ > F₃

therefore the correct answer is 3

You might be interested in

Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha

tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

R_cm = 1 / M ∑ x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

M = m_earth + m_moon

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

M = 5.98 10²⁴ + 7.35 10²²

M = 6.0535 10₂⁴24 kg

we substitute

R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

R_cm = 4.66 10⁶ m

4 0

2 years ago

According to the law of reflection, the angle of incidence is equal to the angle of

Darina [25.2K]

Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r

Hence the correct answer will be angle of reflection.

3 0

2 years ago

Read 2 more answers

How much work can be done in 30 seconds by a 1000-watt microwave oven?

iris [78.8K]

Half of it can be be at 1000 watt microwave oven!!! I hope it help!!!!!!

7 0

2 years ago

Read 2 more answers

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially

ivanzaharov [21]

Answer:

a) s,f,r b) r c) f

Explanation:

To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged

For the sphere to be in equilibrium the sum of forces is zero

B - W = 0

B = W = mg

Now let's use the concept of density for the body and water

Solid sphere

ρ = m / V

V = 4/3 π r³

m = ρ₀ (4/3 π r³)

W = ρ₀ (4/3 π r³) g

Water (a)

ρ = mₐ / Vₐ

mₐ = ρ Vₐ

B = ρ Vₐ g

Let's replace and simplify

ρ Vₐ g = ρ₀ (4/3 π r³) g

ρ Vf = ρ₀ (4/3 π r³) (1)

For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

r <ro V <V₀ ⇒ ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

Assume the three possibilities

- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)

- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)

- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

b) sphere the same radius, but the density increases.

In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

r = r₀ V = V₀ m <m₀ ρ₁ <ρ₀

The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

8 0

2 years ago

The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where

lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

7 0

2 years ago