Which element is LEAST likely to react with Magnesium?

प्रकाशको आवर्तनको परिभाषा लेख्नुहोस् । कन्भेक्स ऐनाको अगाडि तपाई उभियौ भने तपाईको कस्तो आकृति बन्छ, लेख्नर

allochka39001 [22]

A diagram alright of the

4 0

3 years ago

Consider an electron traveling horizontally in the positive direction, above, near, and parallel to a current-carrying wire. The

allsm [11]

Answer:

(a) The electron will move towards the wire.

The direction of the magnetic fields created by the wire can be found via right-hand rule. If you point your thumb towards the direction of the current, and if you curl your fingers, the direction of your four fingers will give the direction of the magnetic field. In this case, magnetic field is around the wire, and into the page just above the wire, where the electron is located.

$\vec{F} = q\vec{v} \times \vec{B}$

According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.

Since the electron has a negative charge, the direction of the force is towards the wire.

(b) The proton will veer to the right.

The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.

6 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

What is transferred to an object when is done A:energy B:force C:effort D:resistance

liberstina [14]

B force i belive becuase when is done it force

6 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

crimeas [40]

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$