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Aliun [14]
3 years ago
15

A 4.1 object is lifted to a height of 4.5m above the surface of Earth.What is the potential energy of the object due to gravity?

​
Physics
1 answer:
liberstina [14]3 years ago
5 0

Answer:

P = 180.81 J

Explanation:

Given that,

Mass of a object, m = 4.1 kg

It is lifted to a height of 4.5 m

We need to find the potential energy of the object due to gravity. It is given by the formula as follows :

P = mgh Where g is acceleration due to gravity

P = 4.1 kg × 9.8 m/s² × 4.5 m

P = 180.81 J

Hence, the potential energy is 180.81 J.

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Answer: I actually need the same answer

Explanation:

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A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
3 years ago
Which of the following would be a good question that could be scientifically investigated?
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Do cherry popsicles freeze slower than orange popsicles
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3 years ago
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Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

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Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

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then the yield YP of the product is

YP= mP / 100 kg =  46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

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3 0
3 years ago
A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa
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Answer:

-5 V

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vb−va

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where:

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q=3.2\cdot 10^{-19} C is the charge of the particle

\Delta V =V_b-V_a is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V

8 0
3 years ago
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