After the cart bounces, how far does it roll back up the ramp?

What type of radioactive decay is shown in this equation?

svp [43]

There is no <span>radioactive decay</span>

6 0

3 years ago

Katrina is a teen from Australia who participates in curling, a sport in which players slide stones across the ice toward a targ

goldenfox [79]

Personally i would say advertising but I can also see where it could be globalization because the Olympics is about different countries coming together and having friendly competition.

6 0

3 years ago

Read 2 more answers

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

Bas_tet [7]

Answer:

W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

F = G m₁ M / r²

W = ∫ F. dr

W = G m₁ M ∫ dr / r²

we integrate

W = G m₁ M (-1 / r)

We evaluate between the limits, lower r = R_ Moon and r = ∞

W = -G m₁ M (1 /∞ - 1 / R_moon)

W = G m1 M / r_moon

Body weight is

W = mg

m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

m = 1000/32 = 165 / g_moon

g_moom = 165 32/1000

.g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system

W_capsule = 1000 pound (1 kg / 2.20 pounds)

W_capsule = 454 N

W = m_capsule g

m_capsule = W / g

m = 454 /9.8

m_capsule = 46,327 kg

Let's calculate

W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶

W = 1,307 10⁶ J

7 0

3 years ago

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

Zigmanuir [339]

Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

8 0

3 years ago

Compare tidal range to spring tide

makkiz [27]

Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

7 0

3 years ago