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dedylja [7]
3 years ago
9

una bala de 20 g choca con un fango como se muestra en la figura y penetra una distancia de 6 cm antes de detenerse. calcule la

fuerza de frenado f si la velocidad de entrada fue de 80m/s​
Physics
1 answer:
aleksklad [387]3 years ago
7 0

Answer:

A 20g bullet collides with a mud as shown in the figure and penetrates a distance of 6cm before stopping. calculate the braking force f if the input speed was 80m/s

Explanation:

Given that,

A bullet of mass

M = 20g = 0.02kg

The bullet is fired into a mud initially at rest

The bullet penetrate a distance of 6cm in the mud

S = 6cm = 0.06m

The input velocity into the mud is 80m/s, this implies that, the initial velocity is 80m/s

U = 80m/s

So, the bullet stop after a distance of 6cm, so the final velocity is 0m/s

V = 0m/s

So, we need to find the braking force

Breaking force (F) is given as

F = ma

Where m is mass and 'a' is deceleration

So, we need to find the deceleration, using equation of motion

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m/s²

So, the negative sign shows that the bullet is decelerating

So, a = 53,333.33 m/s²

So, breaking force

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

The breaking force is 1066.67 N

In spanish

Dado que,

Una bala de masa

M = 20 g = 0.02 kg

La bala se dispara en un lodo inicialmente en reposo

La bala penetra una distancia de 6 cm en el barro.

S = 6 cm = 0.06 m

La velocidad de entrada al lodo es de 80 m / s, esto implica que la velocidad inicial es de 80 m / s

U = 80 m / s

Entonces, la bala se detiene después de una distancia de 6 cm, por lo que la velocidad final es de 0 m / s

V = 0 m / s

Entonces, necesitamos encontrar la fuerza de frenado

La fuerza de ruptura (F) se da como

F = ma

Donde m es masa y 'a' es desaceleración

Entonces, necesitamos encontrar la desaceleración, usando la ecuación de movimiento

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m / s²

Entonces, el signo negativo muestra que la bala se está desacelerando

Entonces, a = 53,333.33 m / s²

Entonces, fuerza de ruptura

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

La fuerza de ruptura es 1066.67 N

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Answer:

Explanation:

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5 0
3 years ago
Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kΩ, and
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Answer:

10042.6 ohm

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f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

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XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2  + (XL - Xc)^2

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Z = 10042.6 ohm

4 0
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Why is the answer B?
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Answer:

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The masses don't have any effect on the outcome: they remain constant.

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4 years ago
A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
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Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
4 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

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Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

5 0
3 years ago
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