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allsm
9 days ago
8

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Physics
1 answer:
Yuliya22 [2.4K]9 days ago
5 0

For this issue, the answer is clarified as the system takes in energy (+). The surroundings contribute 84 KJ of work. Whenever a system is receiving work from its surroundings, the value will be positive. Therefore, it sums to 12.4 KJ + 4.2 = 16.6 KJ.

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A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br
Keith_Richards [2256]
(A) velocity = 2.8 m/s (B) Average force = 1.9536 Newtons.
3 0
6 days ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
serg [2593]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Distance = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

The potential difference is calculated as

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference across the plates amounts to 10000 V

Area is determined by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
1 day ago
Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
Keith_Richards [2256]

The answer is:

V=14m/s

Details are as follows:

According to the problem, we have

The combined mass of A and B is 60kg

A's speed is 2m/s

B's speed is 1m/s

The mass of the bag is 5kg

Typically, the momentum of astronaut A along with the bag is defined by

M_A=(60+5)*2

M_A=130kgm/s

To prevent a collision, astronaut A should maintain a speed that is either equal to or less than astronaut B's speed

Thus, the minimum speed astronaut A should achieve corresponds to that of astronaut B, which is 1

Consequently,

130=(60*1)=(5*v)

V=14m/s

7 0
11 days 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 [2029]

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:

  1. The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
  2. 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
24 days ago
Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
Keith_Richards [2256]

Based on the kinematic formula:

v^2 = v_o^2 + 2ax

We have the following known information:

Acceleration a = 2.55 m/s²

v_0 = 21.8 m/s

v = 0

We want to determine x.

Rearranging the equation, we get:

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

Therefore, the object travels a distance of 93.2 meters.

7 0
1 month ago
Read 2 more answers
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