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3 years ago

Which element is most likely to be shiny?

Physics

2 answers:

mestny [16]3 years ago

5 0

Answer:

Calcium, or as we intellectuals call it, Bone Juice

Explanation:

stellarik [79]3 years ago

4 0

calcium (ca) is most likely to be shiny

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which of the following would be best to do if you were inside a car and a power line fell on the car

NARA [144]

Answer:

Do not move stay in your car and wait for someone from the power company to come and help

Explanation:

Plz vote my answer as the brainiest, i rlly need it! hope this helps!

6 0

2 years ago

What is the difference between kinetic and mechanical energy?

Wittaler [7]

Kinetic energy is energy that is in motion, thats all I remember

8 0

3 years ago

Read 2 more answers

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago

How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?

gtnhenbr [62]

Answer:

W = 47040 J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040 J

So, the required work done is 47040 J.

7 0

2 years ago

A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are

Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, $\theta=30^{\circ}$

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)$

$v = 0$

$Fd\ cos\theta=\dfrac{1}{2}m(u^2) F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N$

The force required to push to stop the car is 288.67 N

3 0

2 years ago