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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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You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr
Nitella [24]

Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

4 0
2 years ago
Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from
Setler [38]

Answer:

given,

mass of copper = 100 g

latent heat of liquid (He) = 2700 J/l

a) change in energy

Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (300 - 4)

Q = 11153.63 J

He required

Q = m L

11153.63 = m × 2700

m = 4.13 kg

b) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (78 - 4)

Q = 2788.41 J

He required

Q = m L

2788.41 = m × 2700

m = 1.033 kg

c) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (20 - 4)

Q = 602.90 J

He required

Q = m L

602.9 = m × 2700

m =0.23 kg

8 0
2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0
2 years ago
Read 2 more answers
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual
kotegsom [21]

Answer:

The amount of heat required is H_t =  1.37 *10^{6} \ J

Explanation:

From the question we are told that

The mass of water is m_w  =  20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g

The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

So

H=   5.7 *10^2 * 4.18 * (36.6 - 3.8)

=> H= 7.8 *10^{4} \  J

Generally the no of mole of sweat present mass of water is

n = \frac{m_w}{Z_s}

Here Z_w is the molar mass of sweat with value

Z_w =  18.015 g/mol

=> n = \frac{5.7 *10^2}{18.015}

=> n = 31.6 \  moles

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

Here L_v is the latent heat of vaporization with value L_v  = 7 *10^{3} J/mol

=> H_v  =  31.6 * 7 *10^{3}

=> H_v  = 1.29 *10^{6} \  J

Generally the overall amount of heat energy required is

H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

4 0
2 years ago
An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac
Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

       = (50 - 0)/4

       = 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0
2 years ago
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