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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

You might be interested in
what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t
kykrilka [37]
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

 
6 0
2 years ago
the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
Andreyy89

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We need to calculate the pressure at this point

Using formula of pressure

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v = velocity

Put the value into the formula

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Hence, The pressure at this point is 0.875 mPa

5 0
2 years ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th
maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0
2 years ago
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
2 years ago
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