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3 years ago

5

Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s

ame length of time, exerting equal force on each, the momentum of the light cart is

1 answer:

Marrrta [24]3 years ago

8 0

F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m a₂ ⇒ a₁ = 2 a₂

since v = at + v₀ with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m

p₁ = p₂

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klasskru [66]

Answer with Explanation:

We are given that

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a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

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3 years ago

A cube-shaped piece of copper has sides of 4cm each and it's density is

iris [78.8K]

Answer:

$64 cm^3$

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

$V=(4\ cm)^3=64\ cm^3$

Surprisingly, no other magnitude is required, thus the answer is:

$\mathbf{64 cm^3}$

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A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.

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Answer:

The wavelength of the light is 562.5 nm

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As per the question:

Order, n = 1

Slit separation, d = $2.5\times 10^{- 6}\ m$

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

$n\lambda = dsin\theta$

n = 1

$\lambda = dsin\theta$ (1)

Also,

For very small angle, $\theta$ :

$sin\theta$ ≈ $tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225$

Using the above value in eqn (1):

$\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm$

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Naddika [18.5K]

Answer:

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Answer:

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