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3 years ago

11

Object a and object b are both in motion when they collide with each other. They then continue in a new direction unaffected by

any other force this is a

1 answer:

Harrizon [31]3 years ago

8 0

This is an elastic collision

bcuz i think they move apart after the collision

sorry if im wrong

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What led up to the disaster? What is the theory on what caused the Tōhoku earthquake

ELEN [110]

Sheer tie a very good wjje

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3 years ago

If in the experiment, m1 is 38 g, m2 is 63 g, and m3 is 58 g, "and m3 remains static, what is the tension in the string connecti

melomori [17]

Let's call a the acceleration of the system. The problem says that the block m3 is static, so the acceleration is zero: a=0.
Calling $T_1$ the tension of the string between m1 and m3, and $T_2$ the tension of the string between m2 and m3, the problem can be solved by writing the following system of equations:
$m_1 g-T_1=m1_a$
$T_1-T_2=m_3 a$
$T_2-m_2g=m_2a$
However, we know that a=0 and the problem asks only for $T_1$ , so we just need to solve the first equation:
$m_1 g -T_1 =0$
and so
$T_1=m_1g=0.038~kg \cdot 9.81~m/s^2 = 0.37~N$

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3 years ago

A child is swinging back and forth on a swing. Changes in her kinetic and potential energy are happening each moment. At which p

mina [271]

Answer:

When the person has swung all the way back (position A), the swing pauses a moment. ... As the swing continues forward, it gradually slows down. Its kinetic energy changes back to potential energy until it reaches the farthest point in its arc (position C).

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3 years ago

A 100.0 gram sample of Polonium-210 is contained for 552 days. How many half-lives occur during this period of time, if the half

Leto [7]

Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

$a_o=a\times e^{-\lambda t}$

$\lambda =\frac{0.693}{t_{1/2}}$

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives and time t

$a_o$ = Initial amount of the reactant.

$\lambda =$ decay constant

$t_{1\2}$ = half life of an isotope

n = number of half lives

We have :

$a_o=100.0 g$

a = ?

t = 552 days

$t_{1/2}=138 days$

$a=100.0 g\times e^{-\frac{0.693}{138}\times 552}$

$a=6.254 g$

$6.254 g=\frac{100.0 g}{2^n}$

$2^n=\frac{100.0 g}{6.254 g}$

n = 4

4 half-lives will occur during this period of time.

4 0

4 years ago

if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a

Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0

3 years ago