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arlik [135]
3 years ago
5

Someone says, “The sun seems brighter because it is bigger than other stars.” What would you say?

Physics
1 answer:
trapecia [35]3 years ago
7 0

Answer: Honestly I would of been like thats cool

Explanation: Hope this helps :)

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Which material is not a fluid? water cork gasoline air
Step2247 [10]

Answer:

Cork

Explanation:

Cork is a solid, other ones are fluid.

3 0
3 years ago
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#86 can an object be moving when its acceleration is zero? can an object be accelerating when its speed is zero?
liraira [26]
Yes, an object<span> that was set in motion in the past by some force, but that is no longer being acted on by a net force, is </span>moving<span> but with </span>zero acceleration<span>, i.e. it is </span>moving<span> at constant velocity.</span>
7 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
4 years ago
What would a force diagram for something WHILE it is being thrown DOWNWARDS look like? <br><br> Ty
Dovator [93]

Answer:

it look the same just to tell you

5 0
3 years ago
A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe
marysya [2.9K]

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration.  So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

7 0
4 years ago
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