Gravity is affected by (2 points)

edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement

Naily [24]

The displacement of Edward in the westerly direction is determined as 338.32 km.

<h3>What is displacement of Edward?</h3>

The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.

The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰

The displacement is the side of the triangle facing 150⁰ = R

R² = a² + b² - 2abcosR

R² = 150² + 200² - (2x 150 x 200)xcos(150)

R² = 62,500 - (-51,961.52)

R² = 114,461.52

R = 338.32 km

Learn more about displacement here: brainly.com/question/321442

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3 0

1 year ago

I really need help please just answer at least one

yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

$v_{f}^{2} = v_{i}^{2}-(2*a*x)\\$

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

$v_{f} =v_{i} - (a*t)$

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

Note: The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

$v_{f}= v_{i}+a*t\\$

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

$v_{f}= v_{i}+a*t\\$

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0

3 years ago

A train is moving at a speed of 50 km/h. How many hours will it take the train to travel 600 kilometers?

EastWind [94]

Answer:

12 hours

Explanation:

600 divided by 50 is 12

6 0

3 years ago

Read 2 more answers

Why did you spin counterclockwise rather than clockwise

rodikova [14]

<span>Because of the orbit of the earth and the sun and the moon. </span>

6 0

3 years ago

Read 2 more answers

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m

Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0

3 years ago