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Alika [10]
2 years ago
12

A car is moving on a straight road in a fixed direction at a constant speed of v = 62 km/h with respect to the road. You wish to

state the kinematic vectors of the motion of the car by using a Cartesian coordinate system whose positive x-axis is pointed in the direction of the motion of the car and the origin is fixed at some point on the road. A. What is the expression for the velocity of the car, using the speed v and the unit vectors i,j, and k?
B. What is the x-component of the position vector, in units of kilometers, at time t=0.015 hr?
Physics
1 answer:
Angelina_Jolie [31]2 years ago
5 0

Answer:

A) V=62 i + 0 j + 0 k

B) X=0.93 km

Explanation:

A) In this situation we are talking about a car moving only in the X- axis, hence the velocity of the car is:

V=62 i + 0 j + 0 k

Where the unit vectors i, j and k represent the components x, y and z in the cartesian plane.

In this sense, each unit vector is defined to have a magnitude of exactly one (1).

B) Velocity is defined as the variation of position in time, if this car is moving only along the x direction we will have:

V=\frac{X}{t}

Clearing the position:

X=V.t

X=(62 km/h)(0.015 h)

X=0.93 km

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Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

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2 years ago
How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?
noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

U=mgh

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

U=(50 kg)(9.8 m/s^2)(15 m)=7350 J

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2 years ago
WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car
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2 years ago
during convection, hot material _____ (expands & sinks, cools & rises, expands & rises, deflates & rises) then m
Lady bird [3.3K]

Answer:

During convection, hot material expands & rises then moves to the side and cools & sinks. this circular pattern is called a convection current.

Explanation:

Convection is one of the three methods of transfer of heat. It occurs only in fluids (liquids or gases).

Convection occurs when there is a source of heat that heats a fluid, such as in a boiling pot of water. The water which is on the bottom of the pot becomes warmer before than the water at the top (because it is closer to the flame), and so it becomes less dense: for this reason, it expands and it becomes rising. On the contrary, the water on top is colder, so it is more dense and starts sinking, replacing the warmer water. As the new part of water gets warmer, it starts rising, and so the process is continuously repeated. This circular current is called convection current.

4 0
2 years ago
Read 2 more answers
. Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device prod
slava [35]

Answer:

4.988kW

Explanation:

According to the question, energy E extracted from the ocean breaker is directly proportional to the intensity I. It can be expressed mathematically as E ∝ I

E = kI where k is the constant of proportionality.

From the formula; k = E/I

This shows that increase in energy extracted will lead to increase in its intensity and vice versa.

If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high

E = 10kW and I = 1.20m

k = 10/1.20

k = 8.33kW/m

To know how much energy E that will be produced when they are 0.600 m high, we will use the same formula

k = E/I where;

k = 8.33kW/m

I = 0.600m

E = kI

E = 8.33 × 0.6

E = 4.998kW

The device will produce energy of 4.998kW when they are 0.600m high.

3 0
2 years ago
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