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1 day ago

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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

2 answers:

inna [987]1 day ago

8 0

4 N is the CORRECT awnser

Keith_Richards [1K]1 day 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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A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

kicyunya [1025]

Answer:

(1.6 × 10¹⁸) /s

Explanation:

The current in a wire is calculated using the formula:

I = (Q/t)

where Q represents the total charge of the electrons in the wire

and t denotes time.

Moreover, Q can be expressed as nq

where n is the number of electrons present in the wire

and q is the charge of a single electron, 1.602 × 10⁻¹⁹ C.

This leads us to the equation I = nq/t.

Thus, (n/t) = (I/q) describes the number of electrons flowing per second through a cross-section.

Inserting our current, I = 260 mA or 0.26 A, along with q = 1.6 × 10⁻¹⁹ C, yields:

(n/t) = (0.26/(1.602×10⁻¹⁹)) = (1.62 × 10¹⁸) /s = (1.60 × 10¹⁸) /s

3 0

3 days ago

A transformer is to be designed to increase the 30 kV-rms output of a generator to the transmission-line voltage of 345 kV-rms.

inna [987]

Answer:

$n_s = 920 \turns$

Explanation:

Given,

Voltage of the primary coil (V_p) = 30 kV-rms

Voltage of the secondary coils (V_s) = 345 kV-rms

number of turns in the primary coil (n_p) = 80 turns

number of turns in the secondary coil (n_s) =?

the ratio of turns between primary and secondary coils

$\dfrac{n_p}{n_s} = \dfrac{V_p}{V_s}$

$\dfrac{n_s}{n_p} = \dfrac{V_s}{V_p}$

$n_s = n_p \dfrac{V_s}{V_p}$

$n_s = 80\times \dfrac{345}{30}$

$n_s = 80\times 11.5$

$n_s = 920 \turns$

The number of turns in the secondary coil is equal to $n_s = 920 \turns$

4 0

1 day ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

Softa [913]

Answer:

176.38 rpm

Explanation:

The proportion of mass for arms and legs is 13%.

For legs and trunk, it's 80% of the total mass.

Additionally, the head accounts for 7% of the total mass.

Overall, the mass of the skater is 74.0 kg.

Each arm length is 70 cm, or 0.7 m.

The skater's height is 1.8 m, and the trunk diameter measures 35 cm, or 0.35 m.

Starting angular momentum is 68 rpm.

We make the following assumptions:

The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
Friction between the skater and the ice is considered negligible.

To analyze her body, we divide it into two parts, treating the arms as spinning rods.

1. Each arm (rod) has a moment of inertia of $\frac{1}{3} mL^{2}$ .

The arms comprise 13% of 74 kg, calculated as 0.13 x 74 = 9.62 kg.

Each arm is then evaluated as 9.62/2 = 4.81 kg.

Let L represent the length of each arm.

Thus,

I = $\frac{1}{3}$ x 4.81 x $0.7^{2}$ = 0.79 kg-m for each arm.

2. The body, treated as a cylinder, has a moment of inertia of $\frac{1}{2} mr^{2}$ .

For the body, the radius r is half of the trunk diameter: r = 0.35/2 = 0.175 m.

The mass of the trunk amounts to (80% + 7%) of 74 kg, which calculates to 0.87 x 74 = 64.38 kg.

Therefore, I = $\frac{1}{2}$ x 64.38 x $0.175^{2}$ = 0.99 kg-m.

Two cases are considered:

case 1: Body spinning with arms extended.

Total moment of inertia equals the combined moments of inertia of both arms and the trunk.

I = (0.79 x 2) + 0.99 = 2.57 kg-m.

The angular momentum is given by Iω.

Here, ω = angular speed = 68.0 rpm = $\frac{2\pi }{60}$ x 68 = 7.12 rad/s.

The angular momentum then becomes 2.57 x 7.12 = 18.29 kg-rad/m-s.

case 2: Arms drawn in alongside the trunk.

The moment of inertia is attributed solely to the trunk. This is 0.91 kg-m.

The angular momentum equals Iω.

= 0.99 x ω = 0.91ω.

By the principle of conservation of angular momentum, the two angular momentum quantities are equal. Therefore,

18.29 = 0.99ω.

Solving gives ω = 18.29/0.99 = 18.47 rad/s.

This leads to 18.47 ÷ $\frac{2\pi }{60}$ = 176.38 rpm.

7 0

3 days ago

A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

kicyunya [1025]

Solution:

The kinetic energy of the clam at an elevation of 5.0 m is 5.19 J and the velocity of the clam at that height is 9.71 m/s.

Explanation:

Throughout its motion, mechanical energy remains constant. We understand that mechanical energy is the summation of potential energy and kinetic energy. Potential energy = $m \times g \times h$ , Kinetic energy = $\frac{1}{2} \times m \times v^{2}$ and Mechanical energy = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Initial kinetic energy is zero. At a height of 9.8 m, the mechanical energy of the clam with a mass of 0.11 kg and g=9.81 $\frac{m}{s^{2}}$ is calculated as follows: 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of the clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ = $5.39+\frac{1}{2} \times m \times v^{2}$ . Given that mechanical energy is conserved, we can state that the mechanical energy of the clam at a height of 9.8 m is equal to that at 5.0 m. The representation is as follows:

10.58 = $5.39+\frac{1}{2} \times m \times v^{2}$ 10.58 – 5.39 = $\frac{1}{2} \times m \times v^{2}$ 5.19 = $\frac{1}{2} \times m \times v^{2}$ the clam's kinetic energy measures 5.19 J.

Lastly, the speed of the clam at 5.0 m is computed; thus, 5.19 = $\frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2}$ 94.36 = $v^{2} \sqrt{94.36}=v \quad v$ = 9.71 m/s. The clam's speed is determined to be 9.71 m/s.

6 0

5 days ago

A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow

Maru [1056]

Answer:

Explanation:

Definitions:

P is an upward acting force.

C represents the vertical contact force from the floor.

W is the crate's weight.

Since P cannot lift the crate, the normal reaction force remains active.

Thus, the forces are related as follows:

P - W + C = 0

Rearranged:

P = W - C

7 0

14 days ago

Read 2 more answers