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3 years ago

Anybody know the answer to this ?

Physics

1 answer:

Firlakuza [10]3 years ago

5 0

Answer:

4 ampere

Explanation:

I= v/R

=12/3

= 4 amp

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A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2

Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass $m_1 = 10 g= 10^{-2} \ kg\\\\$

initial speed $u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\$

block mass $m_2 = 4\ Kg$

initial speed $v_2 =-4.2 \frac{m}{s}$

final speed $v_2= 0$

Let $v_1$ will be the bullet speed after collision:

throughout the consevation the linear moemuntum

$\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\$

$= 320 \frac{m}{s}$

The kinetic energy of the bullet in its emerges from the block

$k=\frac{1}{2} m_1 v_1^2$

$=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J$

3 0

3 years ago

Which of the following statements correctly describe the various applications listed above?

fgiga [73]

Answer:

The answer is "c,d,e, and g".

Explanation:

The correct choices can be defined as follows:

Higher-frequency microwaves aren't used in any of these systems.
Infrared waves aren't seen in each of these technologies.
Its shortest wavelength of all of the technologies listed is the above radiation generated by certain wireless networks.
These devices all produce waves of wavelengths ranging from 0.10 to 10.0 cm.

3 0

3 years ago

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0

3 years ago

PLEASE HELP HDJFJDJDJJCJDJDJFJD

GenaCL600 [577]

Answer:

Explanation:

free

5 0

3 years ago

The number of molecules in a container is tripled and the kelvin temperature doubled. the volume remains unchanged. the new pres

Nonamiya [84]

N2 = 3*n1
T2 = 2*T1
V1 = V2

(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1

Since P2 is 6P1, it is 6 times greater than original pressure

3 0

3 years ago