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ale4655 [162]
2 years ago
9

(2.) Jerry first walks 5m to East, then 6m to N 60 E, then, she walks 16m to N30'W, 3 finally turns he face and moves 3V3m to so

uth. What is her displacement and direction?​
Physics
1 answer:
Vladimir79 [104]2 years ago
5 0

The displacement is determined to be 5.11 m after every direction is plotted on the page.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points.

The steps in which Jerry walks;

5m to East

6m to N 60' E,

16m to N30'W,

Turn

3√3 m to the south.

Each and every direction is plotted on the page and obtained that the displacement is found as 5.11 m.

To learn more about displacement refer to the link;

brainly.com/question/10919017

#SPJ1

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Molodets [167]

Answer:

A

Explanation:

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3 years ago
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A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
patriot [66]

The acceleration due to gravity near the surface of the planet is 27.38 m/s².

<h3>Acceleration due to gravity near the surface of the planet</h3>

g = GM/R²

where;

  • G is universal gravitation constant
  • M is mass of the planet
  • R is radius of the planet
  • g is acceleration due to gravity = ?

g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².

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4 0
2 years ago
What is the net magnetic flux through any closed surface?
Paul [167]

Answer:

The net magnetic flux through any closed surface must always be zero.

3 0
3 years ago
A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output
Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 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
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