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Butoxors [25]
3 years ago
8

Paraffin wax is used to make candlestrue or false​

Physics
1 answer:
tangare [24]3 years ago
7 0

Answer:

true

Explanation:

PLS MAKE ME AS BRAINLIST

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Relative means
Bond [772]

Answer:As a noun relative means a person who is connected with others by blood.

7 0
2 years ago
Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a
timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

a_2 = 2.12 m/s^2

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

F = 0 = m_1a_1 + m_2a_2

now we have

m_1 = 0.200 kg

m_2 = 0.250 kg

a_1 = 2.65 m/s^2

now we have

0.200(2.65) + 0.250a_2 = 0

so we have

a_2 = 2.12 m/s^2

8 0
3 years ago
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
3 years ago
A comparison of a machine’s work output with its work input is called __??__.
Aloiza [94]
Ohhhhh its called a input machine
7 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
lapo4ka [179]

Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

Spring constant of each spring = k

Thus,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

7 0
3 years ago
Read 2 more answers
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