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Damm [24]
3 years ago
5

How does the euglena differ from the paramecium? Question 20 options: Euglena are photosynthetic. Paramecia use flagella to move

. Paramecia have an "eye spot" to detect light. Euglena have both a macronucleus and a micronucleus
Physics
1 answer:
Sedbober [7]3 years ago
3 0

Answer: Euglena are photosynthetic

Explanation: USA test prep

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A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
3 years ago
How do volcanoes above hot spots differ from volcanoes above subduction zones?
BARSIC [14]
C. Volcanoes above subduction zones have lava that is not basaltic.
3 0
3 years ago
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9. From the densities below, decide whether each substance will sink or float in sea water. Sea water
tekilochka [14]

Answer:

Gasoline will float

Asphalt will sink

Cork will float

Explanation:

Simply compare the value of each object's density to that of the sea water (1.025 g/ml). If the density of the object is less than that of the water, the object will float due to the buoyance force.

Contrarily, if the density of the object is larger than that of sea water, the object will sink.

Gasoline, with density 0.66 g/ml which is less than that of sea water, will float.

Gasoline, with density 1.2 g/ml which is more than that of sea water, will sink.

Cork, with density 0.26 g/ml which is less than that of sea water, will float.

8 0
3 years ago
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Natali [406]
A change in internal energy causes work to be done and heat to flow into the system
6 0
3 years ago
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A 200 kg boulder is 1000m above the ground. What is the speed of the object right before it hits the ground?
dem82 [27]

Answer:

140.07m/s

Explanation:

we use the equation for the final speed:

v_{f}^2=v_{0}^2+2gh

where v_{f} is the final speed, v_{0} is the initial speed, g is the gravitational acceleration (g=9.81m/s^2) and h is the height.

In this case we don't have an initial velocity indicated so: v_{0}=0, and we are told that the boulder is at a height: h=1,000m.

We substitute this values into the equation:

v_{f}^2=2gh\\v_{f}^2=2(9.81m/s^2)(1,000m)\\v_{f} ^2=19,620m^2/s^2\\\\v_{f}=\sqrt{19,620m^2/s^2} \\v_{f}=140.07m/s

the speed of the object right before it hits the ground is 140.07m/s

6 0
3 years ago
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