Which of the following is the unit of momentum<br><br>a) ms-1 b) Kg ms-1<br><br>c) Ns d) Kgms-2

What form of energy transforms to what other form when you turn on a steam iron?

timama [110]

Electrical energy to thermal energy

4 0

3 years ago

A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

postnew [5]

Answer:

r=15.53 nm

$F=9.57\times 10^{-13}N$

Explanation:

Lets take electron is in between iron and uranium

Charge on electron $q_1= -1.602\times 10^{-19}C$

Charge on iron $q_2= 2\times 1.602\times 10^{-19}C$

Charge on uranium $q_3= 1.602\times 10^{-19}C$

We know that force between two charge

$F=K\dfrac{q_1 q_2}{r^2}$

$K=9\times 10^9\dfrac{N-m^2}{c^2}$

For equilibrium force between electron and iron should be force between electron and uranium

Lets take distance between electron and uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

$K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}$

$K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}$

$q_2= 2\timesq_1,q_3=q_1$

$\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}$

So r=15.53 nm

So force

$F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}$

$F=9.57\times 10^{-13}N$

7 0

2 years ago

When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the c

vladimir1956 [14]

Answer:Time constant gets doubled

Explanation:

Given

L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by

$i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ]$

where $i_0$ is maximum current

i=current at any time

$\tau =\frac{L}{R}=time\ constant$

$\tau '=\frac{2L}{R}=2\tau$

thus if inductance is doubled then time constant also gets doubled or twice to its original value.

5 0

3 years ago

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

erik [133]

Answer:

The stress is $\sigma = 1.218*10^{6} \ N/m^2$

Explanation:

From the question we are told that

The diameter of the post is $d = 29 \ cm = 0.29 \ m$

The length is $L = 2.0 \ m$

The weight of the loading mass

Generally the radius of the post is mathematically represented as

$r = \frac{0.29}{2}$

=> $r = 0.145 \ m$

Generally the area of the post is

$A = \pi r^2$

=> $A = 3.14 * 0.145 ^2$

=> $A = 0.066 \ m^2$

Generally the weight exerted by the load is mathematically represented as

$F = m * g$

=> $F = 8200 * 9.8$

=> $F = 80360 \ N$

Generally the stress is mathematically represented as

$\sigma = \frac{F}{A}$

=> $\sigma = \frac{80360 }{0.066}$

=> $\sigma = 1.218*10^{6} \ N/m^2$

7 0

2 years ago

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the

Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

x = Vox t

t = x / vox

t = 7.1 / 340

t = 2.09 10-2 s

In this same time the height of the window fell

Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

Voy = (Y + ½ g t²) / t

Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

Vy = Voy - gt₂

Vy = 0 -g t₂

t₂ = Vy / g

t₂ = 27.7 / 9.8

t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

X = Vox t₂

D = 340 2.83

D = 962.2 m

Y = Voy₂– ½ g t₂²

Y = 0 - ½ g t2

H = Y = - ½ 9.8 2.83 2

H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0

2 years ago

Which of the following is the unit of momentuma) ms-1 b) Kg ms-1c) Ns d) Kgms-2​

Which of the following is the unit of momentuma) ms-1 b) Kg ms-1c) Ns d) Kgms-2