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3 years ago

Can someone answer these physics questions

Physics

1 answer:

USPshnik [31]3 years ago

6 0

1. The object moves -25 meters in a period of 3 seconds, so its average velocity is (-25 m)/(3 s) = -25/3 m/s ≈ -8.3 m/s

2. Ms. Patil's average velocity is the displacement over the period divided by the period length: (8 mi)/(0.25 h) = 32 mi/h

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10 PTS!

spin [16.1K]

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3 years ago

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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

7 0

2 years ago

Which of the following changes would decrease the coefficient of friction needed for this ride?

posledela

Friction occurs between two contacting surfaces. The coefficient of friction is very much dependent on the roughness of these surfaces. Some of the many ways in which the coefficient can be lessened or decreased are to lubricate the surface or make it shiny by eliminating the spikes which caused the roughness.

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2 years ago

How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

Natasha_Volkova [10]

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

7 0

3 years ago

Why are most objects not magnetic?

Vedmedyk [2.9K]

Because there’s like no metal stuff idk

6 0

3 years ago

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