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morpeh [17]
3 years ago
13

what logging method greatest impact on forest ecosystems

Physics
1 answer:
lora16 [44]3 years ago
3 0
Are there options to choose e
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A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
ivolga24 [154]

Answer:

Explanation:

Given

mass of rock m=40\ kg

Elevation of Rock h=10\ m

Distance traveled by rock with time

h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

5=0\times t+\frac{1}{2}(9.8)(t^2)

t^2=\frac{10}{9.8}

t=1.004\approx 1\ s

velocity at this time

v=u+at

v=0+9.8\times 1.004

v=9.83\ m/s

6 0
3 years ago
_________ liquids and gases exert a buoyant force.
Eduardwww [97]

Answer:

C. Both

Explanation:

5 0
3 years ago
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Heat can be transferred in several ways. Which of the following is an example of conduction? (DOK 2) A The sun is shining on a m
forsale [732]

Answer:

Explanation:

G yvyvybybybububunununu

8 0
3 years ago
what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of
klio [65]
Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.
6 0
3 years ago
An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

5 0
3 years ago
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