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kondor19780726 [428]
3 years ago
10

Write answers with significant figures:a) 17.35 g +8.498 g​

Physics
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

25.9 g

Explanation:

= 17.35

8.498

________+

= 25.848 g = 25.85 g = 25.9 g

*so sorry if wrong

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A perfectly elastic collision conserves momentum and kinetic energy..

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A rocket initially at rest acceleration at a rate of 99.0meter/second^2. Calculate the distance covered by the rocket if it atta
Helga [31]

Distance traveled by the rocket= 1002 m

Explanation:

initial velocity= Vi=0

final velocity= V=445 m/s

time=4.5 s

Acceleration=a=99 m/s²

using kinematic equation

h= Vi t + 1/2 at²

h= 0(4.5) + 1/2 (99) (4.5)²

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Thus the distance traveled by the rocket= 1002 m

6 0
4 years ago
A hot air balloon traveled 3 hours at a speed of of 450 km/hr. What distance did it travel?
erastova [34]

Answer:

it traveled 1350 kilometers or 839 miles

Explanation:

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3 0
3 years ago
A metal cube, 2.00cm on each side, has a density of 6600 kg/m3. find its apparent weight when it is totally submerged in water.
LekaFEV [45]
Answer: 0.439488 N

Explanation:


The apparent weight of the metal is computed as

apparent weight = weight - weight of the displaced fluid 

To compute the weight of the metal, we use the following formula:

\text{weight} = \rho gV

where 

\rho = \text{density of the metal = 6,600 kg/m}^3
g = \text{gravitational acceleration = 9.81 m/s}^2
V = \text{volume of the metal}

Note that the volume is unknown but we can compute this because the metal is a cube with edge = 2 cm = 0.02 m. So, the volume of the metal is given by

\text{Volume} = \text{edge}^3
\\ = (0.02)^3
\\ \boxed{\text{Volume = 0.000008 m} ^3}

Thus, the weight of the metal is computed as

\text{weight} = \rho gV
\\ = (\text{6,600 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.517968 kg \(\cdot \) m/s}^2
\\ \boxed{\text{weight of the metal} = \text{0.517968 N}}

Next, we compute the displaced weight or buoyancy, which has the following formula

\text{weight of the displaced fluid} = \rho' gV'

where

\rho' = \text{density of the fluid (water) = 1,000 kg/m}^3
\\ g = \text{gravitational acceleration = 9.81 m/s}^2
\\ V' = \text{displaced volume}

Note that the displaced volume is equal to the volume of the submerged metal. Since the metal has a volume of \text{0.000008 m} ^3, the displaced volume is \text{0.000008 m} ^3. 

Thus, the weight of the displaced fluid is calculated as 

\text{weight of the displaced fluid} = \rho' gV'
\\ = (\text{1,000 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.07848 kg \(\cdot \) m/s}^2 \\ \boxed{\text{weight of the displaced fluid} = \text{0.07848 N}}

Therefore,

apparent weight of the metal
= weight of the metal - weight of the displaced fluid 
= 0.517968 N - 0.07848 N
apparent weight of the metal = 0.439488 N

3 0
3 years ago
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