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3 years ago

Suppose you walk 21.5 m straight east and then 12.5 m straight north. How far are you from your starting point (in m)

Physics

1 answer:

stira [4]3 years ago

4 0

Answer:

24.9 m

Explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

You walk 21.5 m straight east and then 12.5 m straight north. This forms a right angled triangle with a horizontal distance of 21.5 m, a vertical distance of 12.5 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Therefore using Pythagoras theorem:

x² = 21.5² + 12.5²

x² = 462.25 + 156.25

x² = 618.5

Taking square root of both sides:

√x² = √618.5

x = √618.5

x = 24.9 m = distance between the ending and starting point.

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Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a

Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

$h = \frac{gt^2}{2}$

solving for t

we have

$t = \sqrt{\frac{2h}{g}}$

substituing all value for time t

$t = \sqrt{\frac{2\times 11.98}{9.8}}$

t = 1.56 s

we know that speed is given as

$v = \frac{d}{t}$

$v =\frac{11.98}{1.56}$

v = 7.67 m/s

7 0

3 years ago

Part B

Licemer1 [7]

The number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

<h3>Average daily power consumption</h3>

The average daily power consumption is the amount of electric energy consumed by an individual on a daily rate.

The average daily power consumption of individuals in USA is 12,100 W-hr.

<h3>Converting watts to kilowatts</h3>

E = 12,100 Whr/1000

E = 12.1 kWh

Thus, the number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

Learn more about power here: brainly.com/question/13881533

#SPJ1

5 0

2 years ago

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

4 years ago

A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a

Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

−

⟹

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

−

⟹

−16

⟹

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0

3 years ago

Compare the distance-time graph for a fast-moving object with the distance time graph of a slower moving object.

SSSSS [86.1K]

The distance vs time graph of a faster moving object would have a greater slope than that of a slower moving one. The object would move a greater distance per unit time

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4 years ago