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3 days ago

The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run

ballistics tests on the pistol at the firing range. If a 0.050-kg bullet were fired from the handgun with a speed of 400 m/s and it traveled 0.080 m into the target before coming to rest, what force did the bullet exert on the target?

Physics

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A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed

Ostrovityanka [3204]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

a= $\frac{30-13-23*sin(45)}{3.1}$

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0

3 months ago

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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is $\sqrt{725}$ ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Now, let's calculate the derivative of distance in relation to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

1 month ago

During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)

Softa [3030]

T= 24.5 feet per second. That’s the speed it attains just before hitting the ground.

7 0

1 month ago

Read 2 more answers