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sukhopar [10]
4 years ago
14

A pilot is flying from DFW to NYC. The air traffic controller directs him to fly 69° Northeast at 243 m/s. What is the y compone

nt of his velocity?
Physics
1 answer:
kolezko [41]4 years ago
8 0

Answer:

v_{y} = 227 m/s

Explanation:

Assumptions: 0° is true North, and 90° is east (along the x-axis).

To solve this problem we must use the expression:

v_{y} = v_{0}sin(\theta)

Where v_{y} is the velocity in the y-direction (East), v_{0} is the total velocity in the direction which the aircraft is travelling, and \theta is the direction the aircraft is travelling (angle from the y-axis).

Using the equation above, we obtain the y-component of velocity

v_{y} = (243)sin(69) = 226.86... m/s which is rounded to 227 m/s (due to the number of significant figures in the question).

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Add these measurements, using significant digit rules:
tia_tia [17]

Here we have to add the two measurements given in the question

The measurement values are given as 1.0090 cm and 0.02 cm.we have to  add them on the basis of significant figure rules.

As per the addition rule in terms of significant figures

1-First we have to select the number of significant digits after the decimal point of each quantity.

2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.

3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division

Now we have to add these two quantities as per the above rule-

         1.0090 cm +0.02 cm

         =1.0290 cm

Here the result  will follow 0.02 which has minimum number of significant figures after the decimal points.

Hence we have to round off the number from 9 of 1.0290

As 9 is  greater than 5 ,so he actual result will be 1.03 cm

       

3 0
3 years ago
how much energy is needed to raise the temperature of 2kg of copper from 0°c to 10°c. the specific heat capacity of copper is 38
kogti [31]

Explanation:

H= mass× specific heat capacity×change in temperature

=2×380×(10-0)

=2×380×10

=7600Joules

4 0
3 years ago
Match the term with its description.
Mrac [35]

A=Ion

B=pH

C=Acid

D=Alkaline

6 0
3 years ago
A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co
Shkiper50 [21]

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

  • K = 9 * 10^9 N m^2 / c^2
  • E = 45 N/C
  • r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
  • Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C           Combine

2*k*q / 0.0375^2 = 45 N/C                                  Divide by 2

kq /(0.0375^2) = 22.5 N/C                                   Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2                                           Find d^2

kq = 22.5 * 0.001406                                            Combine

kq = 0.03164 N/C * m^2                                        Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.


7 0
4 years ago
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