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3 years ago

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Find the length of the third side of the triangle.

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

v

Image result for What is the length of the hypotenuse of the triangle below

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.Nov 23, 2016

Explanation:

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A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?

aksik [14]

ANSWER

C. $F=4.5 \times10^3$ . newtons

EXPLANATION

According to Newton's second law,

$F_{net}=ma$ , where

$m=1.00\times 10^3kg$ is the mass measured in kilograms.

and

$a=4.5ms^{2}$ is the acceleration in metres per second square.

We substitute these values to obtain,

$F=1.00\times10^3 \times 4.5$ .

We rearrange to get,

$F=1.00\times4.5 \times10^3$ .

We multiply out the first two numbers and leave our answer in standard form to get,

$F=4.5 \times10^3 N$ .

The correct answer is C

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3 years ago

the magnitude of the normal force acting on a person with mass of 70 kg standing at rest on the flat ground would be ?

Mekhanik [1.2K]

Answer:

$f = mg \\ = 70 \times 9.8 = |f|$

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3 years ago

If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 V, the current flowing through the circ

AnnyKZ [126]

The formula for calculating the current flowing in a circuit is:
I = V /R
Where:
I = current = ?
V = voltage = 60 V
R = resistance = 12 Ohms
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3 years ago

Read 2 more answers

a golfer is on the edge of a 12.5m bluff overlooking the 18th hole which is located 67.1m from the base of the bluff. she launch

Alja [10]

Answer: $26.359 m/s$

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ (2)

$V=V_{o}-gt$ (3)

Where:

$x=67.1 m$ is the horizontal distance traveled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

$t$ is the time

$y=0 m$ is the final height of the ball

$y_{o}=12.5 m$ is the initial height of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity

$V$ is the final velocity of the ball

Let's begin by finding $t$ from (2):

$t=\sqrt{\frac{2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$ (5)

$t=1.597 s$ (6)

Substituting (6) in (1):

$67.1 m=V_{o} cos(0\°) 1.597 s$ (7)

Finding $V_{o}$ :

$V_{o}=42.01 m/s$ (8)

Substituting $V_{o}$ in (3):

$V=42.01 m/s-(9.8 m/s^{2})(1.597 s)$ (9)

Finally:

$V=26.359 m/s$

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3 years ago

What does a reference point provide

vfiekz [6]

To check progress and direction

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3 years ago