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lora16 [44]
3 years ago
6

a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis

tence offered by the wall
Physics
1 answer:
jok3333 [9.3K]3 years ago
8 0
Mass, m = 4g = 0.004 kg
Velocity, =  50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.

Kenetic Energy of bullet = Resistance offered by the wall.

1/2 mv²                        =  Resistance Force * Distance

(1/2) * 0.04 * 0.5 * 0.5 =  F * 0.1

0.5 * 0.04 * 0.5 * 0.5 =  F * 0.1

0.5 * 0.04 * 0.5 * 0.5/0.1 = F

0.05 = F

Therefore, Resistance offered by the wall = 0.05 N
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The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an
Readme [11.4K]
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
F=8.99\cdot10^{9} Nm^2C^{-2}  \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
7 0
3 years ago
A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.
V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar,  F = 534 N

Co-efficient of kinetic friction, \mu_k = 0.3

Now,

The force against the kinetic friction is given as:

f = \mu_k N = u_k Mg

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

f = 0.3\times108\times9.8

or

f = 317.52N

Now, the net force on to the metal safe is

F_{Net}= F-f

Substituting the values in the equation we get

 F_{Net}= 534N-317.52N

or

F_{Net}= 216.48

also,

 

F_{Net}= M\timesacceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=\frac{F_{Net}}{M}

or

 acceleration of the safe=\frac{216.48}{108}

or

 

acceleration of the safe=2.00 m/s^2

Hence, the acceleration of the metal safe will be  2.00 m/s²

3 0
2 years ago
____ have more energy than ____. A. Radio waves; microwaves B. Ultraviolet rays; X-Rays C. Gamma rays; infrared rays D. Radio wa
Aloiza [94]
C. Gamma rays have more energy than infrared rays
6 0
2 years ago
When the distance between two interacting objects doubles, the gravitational force is
Umnica [9.8K]

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

F=GMm/r^2

here, r is the distance between the objects.

Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

5 0
3 years ago
Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by
OlgaM077 [116]

Explanation:

Area of ring \ 2{\pi} a d a

Charge of on ring d q=-(\ 2{\pi} a d a)

Charge on disk

Q=-\left(\pi R^{2}\right)

\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}

Note: Refer the image attached

8 0
3 years ago
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