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Sergio039 [100]
2 years ago
15

I need the answer please help

Physics
2 answers:
Sergeeva-Olga [200]2 years ago
8 0

Answer:

Option A  

Inertia = the resistance an object has to a change in its state of motion.

timurjin [86]2 years ago
3 0

Answer:

A) Inertia

Explanation:

Inertia is why objects tend to resist changes in their motion. Like a rolling ball will keep rolling unless we try to add friction to it, which then would stop the ball.

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Two balls of mass 0.09 kg hang on strings attached to the same point on the ceiling. The balls are given charges Q that cause th
telo118 [61]

Answer:

Q = 6.33μC

Explanation:

To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0

M: mass of the ball = 0.09kg

T: tension of the string

F_e: electric force between charges

angle = 45°

The electric force is given by:

F_e=k\frac{Q^2}{r^2}

Q: charge of the balls

r: distance between balls = 2m

You divide both equation in order to eliminate the tension T:

tan\theta=\frac{F_e}{Mg}=k\frac{Q^2}{Mgr^2}

By doing Q the subject of the formula and replacing you obtain:

Q=\sqrt{\frac{tan\theta Mgr^2}{k}}=\sqrt{\frac{tan45\°(0.09kg)(2m)^2}{(8.89*10^{9}Nm^2/C^2)}}=6.33*10^{-6}C=6.33\mu C

hence, the charge of the balls is 6.33μC

4 0
3 years ago
Brainlist nd 20 points
nadezda [96]

Answer:

I'm not sure but I think it's 35-39

4 0
3 years ago
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The slope of a linear velocity-time graph tells us the _____ of the object.
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Speed of the object !!!!
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The magnetic strips on credit cards can become demagnetized. What type of magnet are they?
Damm [24]
I think you're fishing for "temporary magnet" or something like that,
but I don't agree with it. 

Credit card strips, refrigerator magnets, recording tape, bar magnets,
and big heavy horseshoe magnets are permanent magnets ... you don't
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magnetic. 

But that doesn't mean that they stay magnetic no matter WHAT you do
to them.  They can be DEmagnetized by being heated, dropped on the
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4 0
3 years ago
Read 2 more answers
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
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