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2 years ago

13

A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit

y 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.

1 answer:

Dmitry [639]2 years ago

6 0

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

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problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi

Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

$g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\ x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m$

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

$t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s$

For T2

$x_{l}=Vbt+(1/2)gt_{2}^{2}\\ as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s$

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0

2 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

2 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

Copper is a metal commonly used for electrical wiring in houses. Which metal conducts electricity better than copper at 20°C?

jek_recluse [69]

The correct answer is silver metal.
Conductivity measures the ability of a material or a metal to transmit energy, conductivity may be electrical, thermal or acoustical conductivity. The most electrical conductive element is silver, followed by copper and Gold. Silver also has the highest thermal conductivity as compare to any element. However, copper and Gold are more often used in electrical applications since copper is less expensive and gold has a much higher corrosion resistance.

6 0

2 years ago

1. The property of a material to absorb liquid like water is called

frez [133]

Answer:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

Explanation:

A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.

3 0

2 years ago