All categories

3 years ago

The force of attraction or repulsion between charged objects depends

Physics

2 answers:

otez555 [7]3 years ago

5 0

<span>Force of attraction or repulsion between charged objects, depends on polarity.</span>

alexandr402 [8]3 years ago

4 0

The force of attraction and repulsion depends upon the amount of charge, polarity and distance between those objects.

You might be interested in

Plsssss HELLPPP ILL MARK BRAINLIESTTTTTT PROMISE

11111nata11111 [884]

Answer:

Nuclear Reaction

Explanation:

Technically fusion and fission should be included, but a nuclear reaction includes both fusion and fission, so it must be the answer.

6 0

4 years ago

Vector A has magnitude 13.0 m and vector B has magnitude 16.0 m. The scalar product A.B is 116 m2. What is the magnitude of the

Elan Coil [88]

Answer:

$\text{Vector product}=172.66\ m^2$

Explanation:

Given that,

The magnitude of vector A, $|A|=13\ m$

The magnitude of vector B, $|B|=16\ m$

Scalar product of A and B, $A{\cdot} B=116\ m^2$

The formula for the scalar product is given by :

$A{\cdot} B=|A||B|\cos\theta$

Where, $\theta$ is the angle between A and B.

$\cos\theta=\dfrac{116}{13\times 16}\\\\\theta=\cos^{-1}\left(0.5576\right)\\\\\theta=56.11^{\circ}$

The formula for the vector product is given by :

$A\times B=|A||B|\sin\theta\\\\=13\times 16\times \sin56.11\\\\=172.66\ m^2$

So, the vector product between these two vectors is $172.66\ m^2$ .

6 0

3 years ago

You have finished running a marathon and you are relaxing and your body is

Nitella [24]

I believe the correct answer is C good luck!

3 0

3 years ago

In the speed-time graph for a moving object shown here, the part which

Nikitich [7]

Answer:

I think the answer is ST. Option B

4 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago