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2 years ago

I have a scientific argument due tomorrow, and I need help ASAP.

Physics

1 answer:

yKpoI14uk [10]2 years ago

6 0

Answer:

can you show the designs please

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A sailboat travels a distance of 600 m in 40 seconds. What speed is it going?

Gekata [30.6K]

Answer:

15 miles /seconds

Explanation:

Distance = 600m

Time = 40 seconds

Speed=?

$speed = \frac{distance}{time} \\ speed = \frac{600}{40}$

Simplify

$\frac{600}{40} = \frac{60}{4} \\ = 15$

7 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t

Andrej [43]

Answer:

ratio = 1 : 4.5

Explanation:

If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:

$F_1=\frac{Gm_1m_2}{r^2}$

The force F₂:

$F_2=\frac{Gm_1(2m_2)}{(3r)^2}$

The ratio:

$\frac{F_2}{F_1}=\frac{2}{9}$

8 0

3 years ago

Does altitude has an effect on weight? PLEASE HELP!

Studentka2010 [4]

Answer:

Yes

Explanation:

The farther something is from the center of mass of an object such as a planet, the lower the gravitational force between them

F = GMm/d²

4 0

3 years ago

a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra

Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= $\frac{v}{f}$

λ= $\frac{30000}{90500}$

λ= 0.311m

5 0

2 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago