Ask question

Ask question

All categories

3 years ago

12

Galileo started modern science because he started using ? Rather than speculation

1 answer:

Eva8 [605]3 years ago

4 0

By using the telescope

You might be interested in

Can a vector be shorter than one of its components

Sauron [17]

It can never be shorter than a component - magnitude of avector is the square root of the sum of the components squared, and a square function never produces a negative number. However, it can be the same size as its component, if that component is the only one

3 0

3 years ago

A hiker travels south along a straight line path for 1.5 hours with an average velocity of 0.75 km/hr, then travels south for 2.

ANTONII [103]

Distance travelled in south direction= 1.5hr*0.75km/hr= 1.125km
Distance travlled in north direction= 0.90*2.5=2.25
Net displacement = 2.25-1.125= 1.125 to the north

7 0

3 years ago

Not a question, but I need help on the last two questions I posted, preferably no links, please...

Andrei [34K]

Answer:

ok...

Explanation:

7 0

3 years ago

Read 2 more answers

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago

If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?

lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

$P=\dfrac{W}{t}$

Work done is equal to the change in kinetic energy of the sprinter.

$P=\dfrac{\dfrac{1}{2}mv^2}{t}$

$P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}$

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

3 0

3 years ago