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s344n2d4d5 [400]
3 years ago
12

An automobile accelerates 1.77 m/s2 over 6.00 s to reach freeway speed at the end of an

Physics
1 answer:
GaryK [48]3 years ago
6 0

Answer:

Startinfg speed is 13.82 m/s

Explanation:

Use equation for realtion between start and final speed :

Vf=Vs+a*t

Vf-final speed

Vs-start speed

Vf=24.44m/s

a=1.77m/s²(acceleration)

t=6.00s(Time)

Vf=Vs+a*t

Vs=a*t-Vf

Vs=1.77m/s²*6s-24.44m/s

Vs=-13.82m/s

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3 years ago
Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

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3 years ago
C. length and mass
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4 years ago
Read 2 more answers
A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight
photoshop1234 [79]

Answer:

t=1.9 sec

Explanation:

From the question we are told that:

Height h=28m

Time t=3s

Generally the Newton's equation for Initial velocity upward is mathematically given by

 s=ut+\frtac{1}{2}at^2

 28=3u-\frac{1}{2}*9.8*3^2

 u=24.03m/s

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 v=\sqrt{24.03-2*9.8*28}

 v=5.35m/s and -5.35m/s

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 v=u+at

 5.35=24.03-9.8t

 t=\frac{28.03-5.35}{9.8}

 t=1.9 sec

4 0
3 years ago
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
3 years ago
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