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tester [92]
3 years ago
6

Explain How can you tell that river water is suspension ?

Physics
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

Streams need vitality to move material, and levels of vitality change as the waterway moves from source to mouth.  

At the point when vitality levels are high, huge rocks and stones can be moved. Vitality levels are generally higher close to a waterway's source, when its course is steep and its valley slender. Vitality levels rise considerably higher in the midst of flood.  

At the point when vitality levels are low, just little particles can be moved (assuming any). Vitality levels are most reduced when speed drops as a waterway enters a lake or ocean (at the mouth).

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A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)
STALIN [3.7K]

The car traverses a distance x after time t according to

x=\dfrac12at^2

where a is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity v after 2 s of travel is given by

v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}

which makes C the most likely answer.

3 0
3 years ago
The process that enhances some properties of an object at the expense of other properties is called:
lidiya [134]

Answer:

The answer is: letter a, pop-out effect.

Explanation:

The "pop-out effect" is a phenomenon which allows the person's precognitive processes to detect a<em> visual stimulus that is potentially the most meaningful one</em> in a person's spatial field of attention. The pop-up effect occurs when a person distinguishes one object from the rest.

For example, when a child chooses among pictures in different colors, it is common for the child to point at colored pictures rather than grayscale pictures. This is an example of a pop-out effect. <u>The properties of the colored pictures is more preferred by the child thus, causing him not to choose or mind the grayscale images.</u>

Thus, this explains the answer.

6 0
3 years ago
1.
Rama09 [41]
A :-) a = v^2 by r
Given - radius = 25 m
velocity = 10 m/s
Solution -
a = v^2 by r
a = ( 10 )^2 by 25
a = 100 by 25
( cut 25 and 100 because 25 x 4 = 100 )
a = 4 m/s^2

.:. The centripetal acceleration of the car
= 4 m/s^2.
8 0
3 years ago
Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no f
tankabanditka [31]

Answer:m_1g

Explanation:

Given

Two masses m_1  and m_2  is released and there is tension T in the string

Suppose a is the acceleration of the system

Therefore from Diagram

For m_1

T-m_1g=m_1a

T=m_1(g+a)------1

for m_2 body

m_2g-T=m_2a

T=m_2(g-a)-------2

From above two Equation it is said that Tension is greater than m_1g and less than m_2g

m_1g

4 0
3 years ago
Read 2 more answers
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
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