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Anna11
10 days ago
11

An 8.0-kg history textbook is placed on a 1.25-m high desk. How large is the gravitational potential energy of the textbook-Eart

h system relative to the desktop?
Physics
2 answers:
ValentinkaMS [3.4K]10 days ago
5 0

Answer

The gravitational potential energy of the textbook-Earth system with respect to the desk is 98 N/m

Explanation

Given the following data:

The mass of the history book is 8kg,

The height from the ground is 1.25m,

The gravitational acceleration is 9.8m/s².

Gravitational potential energy can be calculated using the formula

GPE = Fg⋅h

GPE = mgh

where,

m represents the mass of the object (the book),

g denotes gravitational acceleration,

h signifies the height of the object.

Calculating gives GPE = 8(9.8)(1.25)

       = 98 N/m

Thus, the gravitational potential energy of the textbook-Earth system relative to the desk is calculated as 98 N/m

Ostrovityanka [3.2K]10 days ago
4 0

Answer: 98 J

Explanation: Gravitational potential energy is the energy that an object possesses due to its position within a gravitational field, expressed mathematically as:

G.P.E. = m g h

where m is the object's mass, g is gravitational acceleration, and h represents height.

For the book, m = 8.0 kg,

height of the desk where the book is placed, h = 1.25 m,

and g = 9.8 m/s².

Thus, G.P.E = 8.0 kg × 9.8 m/s² × 1.25 m = 98 J.

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Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
kicyunya [3294]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i + b j, where a and b represent constants to solve for.

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

By comparing the coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Thus,

B = 4.81452 i - 18.38908 j..... 4th quadrant

<pTherefore,

cos(Q) = 4.81452 / 12

Q = 66.346 degrees

360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.

5 0
1 month ago
A cave diver enters a long underwater tunnel, when her displacement with respect to the entry point is 20m,she accidentally drop
Maru [3345]

Answer:

3x864/y488bjehdksuwiieirjr

4 0
1 month ago
slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of
serg [3582]

Response: 800N

Clarification:

Provided data:

Ball mass = 0.8kg

Contact duration = 0.05 seconds

Final and initial speed = 25m/s

The average force exerted by the ball on the wall can be calculated using the following relationship:

Force (F) = mass (m) * average acceleration (a)

a= (initial velocity (u) + final velocity (v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Thus,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Hence,

The average force magnitude (F)

F=ma

m = ball mass = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

6 0
1 month ago
How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.
ValentinkaMS [3465]

Answer:

1.2 × 10^27 neutrons

Explanation:

Considering one neutron weighs 1.67 × 10^-27 kg,

the count of neutrons in a 2 kg mass would be computed as:

2 ÷ 1.67 × 10^-27

Hence, there are approximately 1.2 × 10^27 neutrons.

4 0
1 month ago
The expressions for e/m and the relative error of e/m due to all of the parameters measured:
serg [3582]

Answer:

Term 1 = (0.616 × 10⁻⁵)

Term 2 = (7.24 × 10⁻⁵)

Term 3 = (174 × 10⁻⁵)

Term 4 = (317 × 10⁻⁵)

(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) rounded to the correct significant figures.

Explanation:

(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²

average values

Voltage, V = (403 ± 1) V,

σᵥ = 1 V, V = 403 V

Current, I = (2.35 ± 0.01) A

σᵢ = 0.01 A, I = 2.35 A

Radius of coils, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Radius of curvature of electron path, r = (7.1 ± 0.2) cm.

σᵣ = 0.2 cm, r = 7.1 cm

Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)

Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)

Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)

Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)

The relative e/m value is the total of all the computed terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

= (499 × 10⁻⁵) rounded to the required significant figures.

Hope this Helps!!!

6 0
1 month ago
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