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2 years ago

Explain why humans get an electric shock when the touch live wire ?

Physics

1 answer:

erastovalidia [21]2 years ago

5 0

Answer:

look it up

Explanation:

An electrical shock is received when electrical current passes through the body. ... Whenever two wires are at different voltages, current will pass between them if they are connected. Your body can connect the wires if you touch both of them at the same time. Current will pass through your body.

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A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due

Virty [35]

Answer:

Explanation:

Given

First bicycle travels 6.10 km due to east in 0.21 h

Suppose its position vector is $r_1$

$r_1=6.10\hat{i}$

After that it travels 11.30 km at $15^{\circ}$ east of north in 0.560 h

suppose its position vector is $r_2$

$r_{2}=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )$

after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is $r_3$

$r_{3}=6.10\hat{i}$

so position of final position is given by

$r=r_1+r_{2}+r_{3}$

$\vec{r}=15.12\hat{i}+10.91\hat{j}$

$\vec{v_{avg}}=\frac{\vec{r}}{t}$

t=0.21+0.56+0.21=0.98 h

$\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}$

$|v_{avg}|=\sqrt{361.71}=19.01 km/hr$

For direction

$tan\theta =\frac{11.13}{15.42}=0.721$

$\theta =35.791^{\circ}$ w.r.t to x axis

5 0

4 years ago

If =12a andthe distance from each wire to point p is 0.12m, then what is the magnitude of the magnetic force per unit length on

vaieri [72.5K]

The magnitude of the magnetic force per unit length on the top wire is

2×10⁻⁵ N/m

<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>

To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula

F= $\frac{\mu_0 I_f}{2\pi d}$

Here we are given,

$\mu_0$ = magnetic permeability

= 4 $\pi$ ×10⁻⁷ H m⁻¹

If= 12 A

d= distance from each wire to point.

=0.12m

Now we put the known values in the above equation, we get

F= $\frac{\mu_0 I_f}{2\pi d}$

Or, F = $\frac{4\pi \times 10^{-7}\times 12}{2\pi \times 0.12}$

Or, F= 2×10⁻⁵ N/m.

From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.

Learn more about magnetic force:

brainly.com/question/2279150

#SPJ4

7 0

2 years ago

A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much

Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0

3 years ago

Read 2 more answers

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

A grocery shopper tosses a(n) 9.0 kg bag of rice into a stationary 17.4 kg grocery cart. The bag hits the cart with a horizontal

noname [10]

Answer:

$V=1.77m/s$

Explanation:

#Using the conservation of momentum , momentum before equals momentum after( $p=mv)$ .

-Initial speed is 5.2m/s while the cart is at rest. After, the velocity will be of a combined(bag+cart) mass.

Hence:

$9.0\times 5.2+17.4\times 0=(9+17.4)V\\\\V=\frac{9\times5.2}{9+17.4}\\\\=1.77m/s\\$

Hence, the final velocity of the cart and bag is 1.77m/s

7 0

4 years ago

Explain why humans get an electric shock when the touch live wire ? ​

Explain why humans get an electric shock when the touch live wire ?