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Oksanka [162]
3 years ago
14

Maps are examples of ____ models that people might use everyday.

Physics
2 answers:
Anvisha [2.4K]3 years ago
8 0
A. physical
they represent real things except at a much smaller scale so they are relatable to humans.(..)
Eva8 [605]3 years ago
5 0
I think a. is the right answer..
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A box sits at rest on a rough 33° inclined plane. Draw the free-body diagram, showing all the forces
Cloud [144]

(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

<h3>Free body diagram</h3>

The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;

                      /  W2

                    Ф → Ff

                    ↓W1

where;

  • Ff is the frictional force resisting the down motion of the box
  • W1 is the perpendicular component of the box weight = Wcos(33)
  • W2 is the parallel component of the box weight = Wsin(33)

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

Learn more about free body diagram of inclined objects here: brainly.com/question/4176810

8 0
2 years ago
Electric Charge (guided inquiry)
Sedaia [141]

Answer:

It looks like you have some writing to do ,(-:

Explanation:

8 0
3 years ago
A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the colli
Leto [7]

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

The car is initially stationary.  The truck and car stick together after the collision, so they have the same final velocity.  Therefore:

m₁ u₁ = (m₁ + m₂) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²

ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".

3 0
3 years ago
In a mixture of the gases oxygen and helium, which statement is valid: (a) the helium molecules will be moving faster than the o
mixer [17]

Answer:

a

Explanation:

Given:

In a mixture of the gases oxygen and helium, which statement is valid:

(a) the helium molecules will be moving faster than the oxygen molecules, on average

(b) both kinds of molecules will be moving at the same speed

(c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules

(d) the kinetic energy of the helium will exceed that of the oxygen

(e) none of the above

Solution:

- We will use Boltzmann distribution to answer this question. The root mean square speed of molecules of a gas gives the average speed as follows:

                                        V_rms = sqrt ( 3 k T / m )

- Where, k is the Boltzmann constant, T is the temperature and m is the mass of a single molecule of a gas.

- In general, a mixture has a constant equilibrium temperature T_eq.

- So the v_rms is governed by the mass of a single molecule.

- We know that mass of single molecule of Oxygen is higher than that of Helium molecule. Hence, the relation of mass is inversely proportional to square of root mean speed. So the helium molecules will be moving faster than the oxygen molecules.

- Note: The kinetic energy of the mixture remains constant because it is due to the interaction of the molecules within i.e oxygen and helium. Which makes the kinetic energy independent of mass.

                                     E_k = 0.5*m*v_rms^2

                                     E_k = 0.5*m*(3*k*T/ m )

                                    E_k = 0.5*3*k*T

Hence, E_k is only the function of Temperature which we already established to remain constant at equilibrium.

                                   

8 0
3 years ago
Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k
Yakvenalex [24]

Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

B= magnetic field = 6.90×10-5 T.

We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

= 123388.32m^2/s

Then from the expression above

E =B x dA/dt substitute the values of each parameters, we have

E = 6.90 x 10^-5 x 12388.32 V

E =0.855V

Hence, the average induced voltage between the tips of the wings is =0.855V

6 0
2 years ago
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