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2 years ago

I WILL MARK YOU THE BRAINLIEST NO LINKS

Physics

1 answer:

dsp732 years ago

7 0

Sink because bowling balls sink

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Which is NOT a way to stay safe from static electricity?

lidiya [134]

A run though an open field during a thunderstorm is the answer

5 0

2 years ago

Read 2 more answers

What is the gravitational force between mars and Phobos

alina1380 [7]

Answer:

$F=5.16\times 10^{15}\ N$

Explanation:

We have,

Mass of Mars is, $m_M=6.42\times 10^{23}\ kg$

Mass of its moon Phobos, $m_P=1.06\times 10^{16}\ kg$

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

$F=G\dfrac{m_Mm_P}{d^2}$

Plugging all values, we get :

$F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N$

So, the gravitational force is $5.16\times 10^{15}\ N$ .

8 0

3 years ago

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

levacccp [35]

In component form, the displacement vectors become

• 350 m [S] ==> (0, -350) m

• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0

2 years ago

If Star A is twice as far as Star B, and they are identical in all other ways, then the

baherus [9]

Answer:1/4 the brightness of star b

Explanation:

8 0

3 years ago

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$

Speed of object $u=5\ m/s$

Mass of object at rest $M=3\ kg$

Suppose after collision, speed is v

conserving momentum

$\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s$

Initial kinetic energy

$k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J$

Final kinetic energy

$k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J$

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0

2 years ago