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Licemer1
3 months ago
13

A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal

is 0.2 kg, and it is accelerating to the right 2.5 m/s2.
What is the force that the boy is exerting? N
Physics
2 answers:
inna [3.1K]3 months ago
8 0

4 N is the CORRECT awnser

Keith_Richards [3.2K]3 months ago
5 0
I will assume the girl is on the right while the boy is on the left.
The net force represents the total of all forces acting on an object, factoring in negatives.
Let the force from the boy be denoted as b. We’ll apply the formula F = ma.

b + 3.5 = 0.2(2.5)

This reduces to a straightforward algebraic problem. By solving, we find that the boy is applying a force of -3N to the left.
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What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
Yuliya22 [3333]
In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

6 0
3 months ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Yuliya22 [3333]

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k

7 0
2 months ago
While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti
Sav [3153]

Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.

7 0
3 months ago
Read 2 more answers
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
kicyunya [3294]

Response:

Clarification:

The force between two charges, q₁ and q₂ at a distance d is represented by the formula

F = k q₁ q₂ / d²

Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm

k = 1 / (4π x 8.85 x 10⁻¹²)

Substituting the values into the equation

F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 35969.4 x 10⁻³ N.

For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.

Substituting the values into the expression

F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 82729.6 x 10⁻³ N

Both forces direct towards the left (away from the origin, towards the negative x-axis)

Total force = 118699 x 10⁻³

= 118.7 N.

5 0
1 month ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
ValentinkaMS [3465]

Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

Now converting this to radians since both degree and radian outputs are required:

Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

π

5 0
3 months ago
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