All categories

2 years ago

When an object is 43.0 cm from

Physics

1 answer:

umka2103 [35]2 years ago

8 0

Answer:

hetty dredged ugc dkk xd e vm thx ei

You might be interested in

Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object b? (Remember a=F/m)

Vlada [557]

I would choose the option B.

F = ma

a = 75 / 25 = 3 m/s^2

8 0

2 years ago

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Nastasia [14]

Answer:

750 nm

Explanation:

$d$ = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

$D$ = screen distance = 4.8 m

$y$ = position of first bright fringe = $\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m$

$n$ = order = 1

Position of first bright fringe is given as

$y = \frac{nD\lambda }{d}$

$0.002 = \frac{(1)(4.8)\lambda }{0.0018}$

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago

Please help! Answer and explain

evablogger [386]

Answer:

Explanation:

The asteroid belt is the farthest from the sun, so that means it will take the longest to orbit.

3 0

3 years ago

Please help with both questions thank you!

Katen [24]

I think is:

(4)Inner core

(4)different Kingdome different species

hope this help :)

3 0

3 years ago