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leonid [27]
2 years ago
6

A book that weighs 0.35 kilograms is kept on a shelf that’s 2.0 meters above the ground. A picture frame that weighs 0.5 kilogra

ms will have the same gravitational potential energy as the book when it’s raised to a height of
meters.
(Use PE = m × g × h, where g = 9.8 N/kg. )
Physics
2 answers:
galina1969 [7]2 years ago
4 0
After doing the math, I'm almost 100% sure the answer is "...to a height of 1.4 meters." Glad to help! :)
insens350 [35]2 years ago
3 0

Answer:

A picture frame will have the same gravitational potential energy as the book when it’s raised to a height of 1.4 m.

Explanation:

Given that,

Weight of book = 0.35 kg

Height of shelf = 2.0 m

Weight of frame = 0.5 kg

Using formula of gravitational potential energy

Gravitational potential energy = mgh

Where, m = mass

g = acceleration due to gravity

h = height

A picture frame that weighs 0.5 kilograms will have the same gravitational potential energy as the book when it’s raised to a height.

Therefore,

Gravitational potential energy of book = gravitational potential energy of frame

mgh=m'gh'

We substitute the value into equation (I)

0.35\times 9.8\times2.0=0.5\times9.8\times h'

h'=\dfrac{0.35\times2.0}{0.5}

h'=1.4\ m

Hence, A picture frame will have the same gravitational potential energy as the book when it’s raised to a height of 1.4 m.

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A lead fishing weight of mass 0.2 kg is tied to a fishing line that is 0.5 m long. the weight is then whirled in a vertical circ
Mariana [72]
In the movement of the weight in vertical circle, using momentum balance, the largest tension is at the bottom of the circle. This is represented by: 

<span>F = T - m g </span>
<span>T = F + m g 
</span>F (centripetal) = mv^2/r
<span>= m v^2 / r + m g </span>

<span>m v^2 / r = T - m g </span>
<span>T= 0.5m * 100kgm/s^2 / 0.2kg - 9.81m/s^2 * 0.5m </span>
<span>T= 245 m^2/s^2 </span>


7 0
2 years ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0
2 years ago
Read 2 more answers
Which of the following best describes the dependant variable?
weeeeeb [17]
Hello!

The independent variable is the variable deliberately changed.

The dependent variable is the variable that responds to change. So the answer is A.

Hope this helps. Any questions please just ask! Thank you!
5 0
2 years ago
The expressions for e/m and the relative error of e/m due to all of the parameters measured:
bija089 [108]

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) to the appropriate significant figures.

Explanation:

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

mean measurements

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

Coils radius, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Curvature of the electron trajectory, 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 value of the e/m ratio is a sum of all the calculated terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

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

Hope this Helps!!!

6 0
2 years ago
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