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Liula [17]
3 years ago
8

Doug is doing work. Without knowing exactly what he is doing, which can be said about the work he is doing? Energy is transferre

d; Energy is increased; Energy is decreased; Energy is unchanged;
The answer is Energy is transferred! I just took the test and your answer was Energy is decreased and that was not correct!
Physics
2 answers:
NikAS [45]3 years ago
5 0
Energy is transferred
<span />
fenix001 [56]3 years ago
3 0
The answer is Energy is transferred

There is always energy transferred every time we do a certain amount of work (even though we're not sure what we're doing like Doug) . The type of energy transfer that happen would be determined by the type of work he got.

For example, if the work that Doug donig requires physical work, his energy will be decreased. If the work that Doug doing absorb substance from external environment (such as eating foods), his energy willl be increased.

Nonetheless, there wil always be a transfer of energy in every work
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Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle
tangare [24]

Answer:

F=1.159

Explanation:

From the question we are told that:

Mass of pulley M=1kg

Radius r=12cm

Mass of block A M_a=2.1kg

Mass of block B m_b=4.1kg

Spring constant\mu= 358 J/m2

Generally the equation for Torque is mathematically given by

Since \sumF=ma

At mass A

 T_2-f_3=2.1a

At mass B

 4.8-T_1=4.1a

At  Pulley

 R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}

 R(T_1-T_2)=0.55a

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a

 -f_3+4.1=6.75a

 -f_3=6.75a+4.8

Since From above equation

M_{eff}=6.7kg

Therefore

T=2\pi \sqrt{{\frac{M_{eff}}{k}}

T=2\pi \sqrt{{\frac{6.75}{\mu}}

T=0.862s

Generally the equation for frequency is mathematically given by

F=\frac{1}{T} \\F=\frac{1}{0.862}

F=1.159

3 0
3 years ago
the density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
statuscvo [17]
The type of relationship between atmospheric density and altitude is therefore inverse relationship. This means an increase in either will decrease the other factor. Density is mass per unit volume, it is difficult to calculate with only altitude 1,291 kilometers given. Else, we could also use temperature and pressure to solve density but they are not provided.
5 0
3 years ago
Read 2 more answers
On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow
Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

Total heat

=10465 +167000+ 37638

=215103 J

7 0
4 years ago
A 70 kg object is on a table in static equilibrium. What is the support (normal) force the table exerts on the box
sweet-ann [11.9K]
F(n) = mg = 70 x 9.8 = 68.6 N
3 0
4 years ago
The moon Phobos orbits Mars
shepuryov [24]

27.9816 \times 10^{3} s is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

                  \frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant  6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}

Here, given data

M=6.23 \times 10^{23} \mathrm{kg}

R=9.38 \times 10^{6} \mathrm{m}

Substitute the given values, we get T as

      \frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}

      T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}

      T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}

Taking square root, we get

       T=27.9816 \times 10^{3} s

4 0
3 years ago
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