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4 years ago

12

What is the acceleration of a ball traveling horizontally with an initial velocity os 20 meters/second and ,2.0 seconds later, a

velocity of 30 meters/second?

1 answer:

ch4aika [34]4 years ago

5 0

Acceleration is the rate of change of velocity, so can be calculated by:
a = v2 - v1 / t where v2 and v1 are the final and initial velocities of the object respectively and t is the time taken. Therefore the acceleration is:
a = 30 m/s - 20 m/s / 2.0 s = 10 m/s / 2.0 s = 5 m/(s^2)
Hope this helps!

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Electric current in a (circuit) is due to a (positive charge flow or just flow) produced by (sources such as batteries, solar cells, electric machines). It’s opposed by (resistance).. and I don’t understand the large and small.. but it’s a start. Hopefully I helped

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3 years ago

2. A marble is rolling at a velocity of 1.5 m/s with a momentum of 3.0 kg×m/s. What is its mass?

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Answer:

m= 2[kg]

Explanation:

Linear momentum can be calculated as the product of mass by Velocity.

$P=m*v$

where:

P = linear momentum = 3 [kg*m/s]

v = velocity = 1.5 [m/s]

m = mass [kg]

$m=P/v\\m=3/1.5\\m=2[kg]$

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3 years ago

A 400 N steel ball is suspended by a light rope from<br>the ceiling. The tension in the rope is

alina1380 [7]

Answer:

As force = tension in string so tension in string is 400N

Explanation:

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3 years ago

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12Hz. When

anzhelika [568]

Answer:

Explanation:

Given

When $m_1$ mass is attached to the spring its frequency is $f_1=12\ Hz$

when another mass $m_2$ is attached to $m_1$ , frequency changes to $f_2=4\ Hz$

frequency of spring mass system is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

for $m_1$

$f_1=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}$

$12=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}-----1$

for $m_1$ and $m_2$

$f_2=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}$

$4=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}----2$

divide 1 and 2

$3=\sqrt{\frac{m_1+m_2}{m_1}}$

squaring

$9=1+\frac{m_2}{m_1}$

$\frac{m_2}{m_1}=8$

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3 years ago

A radioactive substance decays at an annual rate of 131313 percent. If the initial amount of the substance is 325325325 grams, w

anyanavicka [17]

Answer: $f_{(t)}=325(0.87)^{t}$

Explanation:

If we have an initial amount of a radioactive material or substance, which is $325g$ , and we also are told this amount decays $13\%$ each year, this means each year $87\%$ of the substance remains:

$100\%-13\%=87\%=0.87$

<u />

<u>To understand it better:</u>

Year 1: $325(87\%)=325(0.87)$

Year 2: $325(0.87)(0.87)=325(0.87)^{2}$

Year 3: $325(0.87)(0.87)(0.87)=325(0.87)^{3}$

and so on until year $t$ :

Year t: $325(0.87)^{t}$

Therefore, the function tha best describes this radiation decay situation is:

$f_{(t)}=325(0.87)^{t}$

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3 years ago