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Andre45 [30]
2 years ago
9

Overall force acting on an object.

Physics
1 answer:
arsen [322]2 years ago
3 0

Answer:

Gravity

Explanation:

Gravity is a force that acts upon every single object on Earth.

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half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds
Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %


3 0
3 years ago
6. Show that the weight of an object on the moon is 1/6 its weight on earth.​
mojhsa [17]

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

5 0
2 years ago
A toy gun uses a spring to project a 5.9-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
LekaFEV [45]

Answer:

Explanation:

Stored energy in spring = 1/2 k x² , k is spring constant , x is compression.

= 1/2 x 8 x (5.7 x 10⁻²)²

= 129.96 x 10⁻⁴ J

Energy lost due to friction = force x distance

= .035 x .17

= .00595 J

Energy used in providing kinetic energy to projectile.  

129.96 x 10⁻⁴  - .00595

.012996 - .00595

= .007046 J

So

1/2 m v² = .007046

v² = .007046  x 2 / .0059

= 2.3885

v = 1.545 m /s

8 0
3 years ago
Read 2 more answers
what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific
weqwewe [10]

Answer:

115.2^{\circ}C

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by \Delta T, according to the equation

Q=mC_s \Delta T

where C_s is the specific heat capacity of the substance.

In this problem, we have:

Q=250 \cdot 4.184 =1046 J is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

C_s = 0.175 J/gC is the specific heat capacity of gold

Solving for \Delta T, we find the change in temperature

\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}

And since the final temperature was

T_f = 175^{\circ}

The initial temperature was

T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C

3 0
3 years ago
Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0 kg of ice at 0ºC?
meriva
If you're only melting ice ... turning solid ice at 0°C into liquid water at 0°C ...
then you only need to know t<span>he latent heat of fusion for water</span>. 

That's exactly what it means ... the amount of energy involved when a gram
of H₂O changes in either direction between liquid and solid, with no change
in temperature.



7 0
3 years ago
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