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andreev551 [17]
3 years ago
5

It took 3.5 hours for a train to travel the distance between two cities at a velocity

Physics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

420

Explanation:

420 because 120 x 3 = 360

120/2 = 60 and 360 + 60 = 420

I hope this helps you and please consider giving me brainliest :)

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When you see yourself in a plane mirror, the image is always:
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It’s c “ the same size as you are”
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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
3 years ago
A 70.0 kg person climbs stairs, gaining 3.90 meters in height. Find the work done (in J) to accomplish this task.
Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = 9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J

6 0
2 years ago
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Effectus [21]

Answer:

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2 years ago
How does a steam engine do work?
Viktor [21]
Steam enters a cylinder—- A
4 0
3 years ago
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