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defon
3 years ago
8

Please help with these i dont know how to do them

Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

<em><u>2</u></em><em><u>0</u></em><em><u>.</u></em><em><u>0</u></em><em><u>M</u></em><em><u>/</u></em><em><u>S</u></em><em><u>. </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>HORIZONTAL</u></em><em><u> </u></em><em><u>VELOCITY</u></em><em><u> </u></em><em><u>OF</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>BALL</u></em><em><u> </u></em><em><u>JUST</u></em><em><u> </u></em><em><u>BEFORE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>REA</u></em><em><u>CHES</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>GROUND</u></em>

<em><u>1</u></em><em><u>2</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>SECONDS</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>APPROXIMATE</u></em><em><u> </u></em><em><u>TOTAL</u></em><em><u> </u></em><em><u>TIME</u></em><em><u> </u></em><em><u>REQUIRED</u></em><em><u> </u></em><em><u>FOR</u></em><em><u> </u></em><em><u>THE</u></em><em><u> </u></em><em><u>BALL</u></em>

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I tried the experiment of dropping candy into a liter of soda and caused a big explosion of foamy soda. Is that a physical or Ch
Aleks04 [339]

Answer:

physical change

Explanation:

its a physical change because its still soda it hasn't changed into anything else than what it already is

6 0
3 years ago
Sally and Sam are in a spaceship that comes to within 16,000 km of the asteroid Ceres. Determine the force Sally experiences, in
Diano4ka-milaya [45]

Answer:

  0.017 N

Explanation:

The relevant relation is ...

  F = GMm/r²

where G is the universal gravitational constant, 6.67408 × 10^-11 m^3·kg^-1·s^-2, M and m are the masses of the objects, and r is the distance between them.

__

Filling in the given numbers, we find the force to be ...

  F = (6.67408 × 10^-11 m^3·kg^-1·s^-2)(8.7 × 10^20 kg)(77 kg)/(1.6 × 10^7 m)^2

where m in this expression is the unit "meters".

  F = 6.67408 · 8.7 · 77/2.56 × 10^(-11 +20 -2·7) N ≈ 0.017 N

The asteroid exerts a force of about 0.017 N on Sally.

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<em>Additional comment</em>

That's about 0.000023 times the force of Earth's gravity.

7 0
2 years ago
1.) A projectile is launched at 15 degrees. The landing height is the same as the launch position. At what other angle can the p
allsm [11]
1) The general equations of motion of the projectile on the x and y axis are:
x(t) = v_0 \cos \alpha t
y(t)=v_0 \sin \alpha t -  \frac{1}{2}gt^2
where v0 is the initial velocity, \alpha is the angle with respect to the ground, and g=9.81 m/s^2 is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.

First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of 15^{\circ}. To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
v_0 \sin \alpha t -  \frac{1}{2}gt^2 =0
t( v_0 \sin \alpha -  \frac{1}{2} gt)=0
that has two solutions: t=0 (beginning of the motion) and
t= \frac{2 v_0 \sin \alpha}{g}
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
x_1=v_0 \cos \alpha t = v_0 \cos \alpha ( \frac{2 v_0 \sin \alpha }{g} )= \frac{2 v_0^2}{g} \sin \alpha \cos \alpha
with \alpha=15^{\circ}.

If we call \beta the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
x_2 =  \frac{2 v_0^2}{g} \sin \beta \cos \beta
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
\sin \alpha \cos \alpha = \sin \beta \cos \beta
Now let's remind that \cos \theta= \sin (90^{\circ} -\theta) so that we can rewrite the equation as
\sin \alpha \sin (90^{\circ}-\alpha) = \sin \beta \sin (90^{\circ}-\beta)
and using \alpha=15^{\circ}:
\sin 15^{\circ} \sin (75^{\circ}) = \sin \beta \sin (90^{\circ}-\beta)
and we can see that there are two values of \beta that satisfy the equation: \beta=\alpha=15^{\circ} and \beta=75^{\circ}, which is the solution of our problem.

2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
7 0
3 years ago
8 points
Lunna [17]
The pilot might be correct (I think), because, if the gravity of the planet is strong, then the planet’s gravity will pull the spaceship into its orbit, so the engines don’t need to be on for the ship to get pushed toward the planet.
4 0
3 years ago
Read 2 more answers
If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?
Irina-Kira [14]
The correct answer is B. <span>0.002010812m3. Good Luck! :)</span>
5 0
3 years ago
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