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2 years ago

14

A runner starts from rest and attains a speed of 8.00ft/s after 2.00s. What's the runner's acceleration? (keep the answer in ft/

s^2)

2 answers:

solong [7]2 years ago

8 0

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arlik [135]2 years ago

3 0

Answer:

4 ft/s^2

Explanation:

If the velocity is 8ft/s after 2s then to find the acceleration you would use the formula:

$\frac{v_f - v_{i} {}}{t} =a$

sooooooooo

$\frac{8 ft/s - 0 ft/s }{2 s} = \frac{8 ft/s }{2 s} = 4 ft/s^{2}$

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A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u

skelet666 [1.2K]

Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

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2 years ago

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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

<u>For </u><u>student 1</u><u>:</u>

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

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2 years ago

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Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a

Yuliya22 [10]

Answer:

The value is $q = 3.4 *10^{-6} \ C$

Explanation:

From the question we are told that

The mass of each sphere is $m_1 = m_2 = m = 0.020 \ kg$

The length of the string is $l = 0.18 \ m$

The angle of with the vertical is $\theta = 7^o$

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally the force acting between the forces is mathematically represented as

$F = T cos \theta = \frac{k* q^2}{ r^2}$

=> $T cos \theta = \frac{k* q^2}{ r^2}$

Generally from Pythagoras theorem the radius of the circular curve created by the force is

$r = 2 L sin (\theta )$

=> $r = 2* 0.180 sin (7)$

=> $r = 0.043 \ m$

=> $q = tan \theta * \frac{m * g * r^2 }{k}$

=> $q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}$

=> $q = 3.4 *10^{-6} \ C$

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2 years ago

An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other

Umnica [9.8K]

The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹ N*m/C²

It is given:

F = 7.2 N

d = 0.1 m = 10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?

Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

7.2 = 36 × 10³ * Q2 / 10⁻²

Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶

Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

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2 years ago

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

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2 years ago