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Alenkinab [10]
3 years ago
9

David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady

2.90 m/s^2 at the instant when David passes.
Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?
Physics
1 answer:
rodikova [14]3 years ago
7 0

Answer:

Explanation:

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1

For David u = 28.0 m/s where 'a' is set to nought

S = ut

S = 28t.......2

For Tina consider equation 1

Where acceleration = 2.90m/s^2 and u is set at nought

S = 1/2×2.90 m/s×t^2.......3

Equate 2 and 3

28t = 1.45t^2

Divide through by t

28 = 1.45t

t = 28/1.45

t = 19.31seconds

Now put the value of t into equation 3

S = 1/2×2.90 m/s×t^2.......3

= 1.45×20×20

= 580m

Tina must have driven 580meters before passing David

Considering the equation of linear motion : V^2 = U^2+2as

Where u is set at nought

V^2 = 2as

V^2 = 2×2.9×580

V^2 = 3364

V = √3364

V = 58m/s

Her speed will be 58m/s

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ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,

F=ma

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

F=ma\\ =(9.0kg)(3m/s^2)\\ =27N

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

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3 years ago
A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a
Slav-nsk [51]

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For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

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8 0
3 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.70 m/s. Her husband Bruce suddenly realiz
Katen [24]

Answer:

The velocity is v  =  6.66 \  m/s

Henrietta is at distance s=  18.17 \  m from the under the window

Explanation:

From the question we are told that

The speed of Henrietta is v=  2.70 \ m/s

The height of the window from the ground is h  =  36.5 \  m

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

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

=> t  =  \sqrt{\frac{2 *  36.5 }{9,8} }

=> t  =  2.73 \  s

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

T = 2.73  +  4

=> T = 6.73 \  s

Generally the distance covered by Henrietta before catching her lunch is

s=  v  *  T

=> s=  2.70  * 6.73

=> s=  18.17 \  m

Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

v  =  6.66 \  m/s

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