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garri49 [273]
1 year ago
8

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

ch. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.
Physics
2 answers:
madreJ [45]1 year ago
7 0
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
xxTIMURxx [149]1 year ago
5 0

Answer:

v_f = 6.48 m/s

Explanation:

As we know that her speed while she walk is given as

v_o = 1.4 m/s

now here acceleration is given as

a = 0.20 m/s^2

now the distance moved by her

d = 100 m

now the final speed is given as

v_f^2 - v_0^2 = 2 a d

so we have

v_f^2 - 1.4^2 = 2(0.20)(100)

v_f = 6.48 m/s

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Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

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For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

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E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

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1 year ago
Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe
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Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
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Explanation:

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Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

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1 year ago
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
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Answer:

25.82 m/s

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We are given;

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Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
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