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umka21 [38]
3 years ago
14

you are doing an experiment outside on a sunny day you find the temperature of some sand is 28 degress Celsius you also find the

temperature of some water is 25 degrees Celsius explain the difference in temperatures
Physics
1 answer:
OLga [1]3 years ago
3 0
E the temperature of a substance. Water has a very high specific heat. That means it needs to absorb a lot of energy before its temperature changes. Sand , on the other hand, have lower specific heats. This means that their temperatures change more quickly. When the summer sun shines down on them, they quickly become hot.
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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
A longitudinal wave is a combination of a transverse wave and a surface wave.(true or false)
Lisa [10]

Answer: True

Explanation:

8 0
4 years ago
If love is the answer, then what is the question?
kipiarov [429]

Answer:

what is happiness

Explanation:

7 0
3 years ago
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What is the distance from the earth's center to a point outside the earth where the gravitational acceleration due to the earth
Crazy boy [7]
R2^ 2 / R1 ^2 = g1 / g2 = 38 

<span>R2 = R1 x √38 = 6.1644* R1 </span>

<span>R2 = 6.1644 x 6378 000 = 39316632.5 m</span>
8 0
3 years ago
2. What is the least possible error encountered
AlekseyPX

Answer: b 0.5mm

Explanation:

6 0
3 years ago
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