Can someone please help, ty!!<br> (Will mark brainliest)

Hi I have a question it’s not about the subject but is at the same time what is Physics?

noname [10]

Answer:

the branch of science that is concerned with nature and properties of matter and energy.

Explanation:

a study of the basis of what does what in science.

3 0

2 years ago

50 points!

Tanya [424]

From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.

5 0

3 years ago

Vestobular receptors enable one to balance. <br><br> True or False

frozen [14]

The answer is actually true. i just took the test and i put false and it was wrong

5 0

3 years ago

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Two protons are going up when they enter a magnetic field that points toward you. What happens to the motion of the particles

asambeis [7]

Force on a proton when they are moving inside the magnetic field is given by

$F = q(v x B)$

so here while protons are going upwards we can say its direction of velocity is towards +Y and magnetic field is towards us which is along +Z direction so they will have force due to this motion.

As per above equation of force we can have direction of force which is along +X direction.

So here both proton will bend towards +X direction and start moving in curved path.

8 0

3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s

3 0

4 years ago

Read 2 more answers

Can someone please help, ty!! (Will mark brainliest)