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grigory [225]
3 years ago
15

Where do most water molds and downy wildews live?

Physics
1 answer:
Leokris [45]3 years ago
5 0
Most water molds and/or downy mildews live in the water or in moist places, These organisms usually grow as tiny fuzz.
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Which will ensure laboratory safety during the experiment? Check all that apply. using beaker tongs to handle the hot beaker rea
Aleks [24]

Answer: • using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use

Explanation:

The options that will ensure laboratory safety during the experiment will be:

• using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use.

We should note that the beaker tongs are simply used in the holding of the beakers that have hot liquids in them. Also, it s vital for the hot plate to be turned off after its use so as to prevent accident.

7 0
3 years ago
Read 2 more answers
while spinning in a centrifuge a 70.0 kg astronaut experiences an acceleration of 5.00 g, or five times the acceleration due to
ollegr [7]

Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

4 0
2 years ago
What force must be used to do 224 Joules of work on an object over a distance of 32 meters?
cestrela7 [59]
7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons
4 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
2 years ago
If an object has 180 J of PE and a mass of .5kg, what is its height?
Nonamiya [84]
PE=mgh
180=0.5*10*h
180=5h
h=180/5=36 m
7 0
2 years ago
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