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3 years ago

How do you get this equation?please help

Physics

1 answer:

Verizon [17]3 years ago

6 0

Answer:

It's a kinematic equation.

Explanation:

x displacement = (velocity_final * time) - (0.5 * acceleration * time * time)

You should be given a value, context(word problem), or diagram.

If you're deriving for a term just isolate the term.

If you're asking why the -(0.5 * acceleration * time * time) is negative instead of positive, would need context.

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What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)

RSB [31]

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

6 0

3 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

3 years ago

The threatened or actual use of physical force to get someone to have sex without consent is called

Nimfa-mama [501]

Wouldn't it be r a p e?

5 0

3 years ago

Read 2 more answers

Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam

Hatshy [7]

Think of it like a graph. You start at the origin which is (0,0). go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6). 1 to the east, ((4,6). Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite.

Hope this helps!

6 0

3 years ago

In a mixture of the gases oxygen and helium, which statement is valid: (a) the helium molecules will be moving faster than the o

mixer [17]

Answer:

Explanation:

Given:

In a mixture of the gases oxygen and helium, which statement is valid:

(a) the helium molecules will be moving faster than the oxygen molecules, on average

(b) both kinds of molecules will be moving at the same speed

(d) the kinetic energy of the helium will exceed that of the oxygen

(e) none of the above

Solution:

- We will use Boltzmann distribution to answer this question. The root mean square speed of molecules of a gas gives the average speed as follows:

V_rms = sqrt ( 3 k T / m )

- Where, k is the Boltzmann constant, T is the temperature and m is the mass of a single molecule of a gas.

- In general, a mixture has a constant equilibrium temperature T_eq.

- So the v_rms is governed by the mass of a single molecule.

- We know that mass of single molecule of Oxygen is higher than that of Helium molecule. Hence, the relation of mass is inversely proportional to square of root mean speed. So the helium molecules will be moving faster than the oxygen molecules.

- Note: The kinetic energy of the mixture remains constant because it is due to the interaction of the molecules within i.e oxygen and helium. Which makes the kinetic energy independent of mass.

E_k = 0.5*m*v_rms^2

E_k = 0.5*m*(3*k*T/ m )

E_k = 0.5*3*k*T

Hence, E_k is only the function of Temperature which we already established to remain constant at equilibrium.

8 0

4 years ago

How do you get this equation?please help​

How do you get this equation?please help