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kati45 [8]
3 years ago
15

What is one basic need that both monkey and a tree have in common​

Physics
2 answers:
vodka [1.7K]3 years ago
6 0

Answer:

water

Explanation:

both a monkey and a tree need water to survive.

Have a great day

aniked [119]3 years ago
3 0
Water both monkey and tree have in common
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What is air temperature meausered with?
vodka [1.7K]

Temperature is usually expressed in degrees Fahrenheit or Celsius. 0 degrees Celsius is equal to 32 degrees Fahrenheit. Room temperature is typically considered 25 degrees Celsius, which is equal to 77 degrees Fahrenheit.

5 0
3 years ago
Which activity is a model of thermal energy transfer by conduction? A. Freezing ice cubes in a freezer B. Using a heat lamp to w
Ne4ueva [31]

Answer:

Hot water rises and cold water sinks is a model of thermal energy transfer by conduction.

3 0
3 years ago
Read 2 more answers
A double slit that is illuminated with coherent light of wavelength 644 nm produces a pattern of bright and dark fringes on a sc
shutvik [7]

Answer:

2.77 cm

Explanation:

d = separation between the slits = 2783 x 10⁻⁹ m

\lambda = wavelength of coherent light = 644 nm = 644 x 10⁻⁹ m

D = Distance of the screen = 6 cm = 0.06 m

y_{n} = Position of nth bright fringe

Position of nth bright fringe is given as

y_{n} = \frac{nD\lambda }{d}  

for n = 2

y_{2} = \frac{nD\lambda }{d}  

y_{2} = \frac{(2)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}

y_{2} = 0.0278 m

for n = 4

y_{4} = \frac{nD\lambda }{d}  

y_{4} = \frac{(4)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}

y_{4} = 0.0555 m

Distance between 4th and 2nd bright fringes is given as

w = y_{4} - y_{2} = 0.0555 - 0.0278 = 0.0277 m

w = 2.77 cm

8 0
3 years ago
A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750
White raven [17]

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force N^> (750 N)  and the tension in the Achilles tendon F^>_{Achilles} (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| N^> | + |F^>_{Achilles} | = | F^>_{Tibia} |

so force exerted on the foot by the tibia will be;

| F^>_{Tibia} | = |N^> | + |F^>_{Achilles} |

so we substitute IN OUR VALUES

| F^>_{Tibia} | = 750 N + 2225 N

| F^>_{Tibia} |  = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

3 0
3 years ago
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
3 years ago
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