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Nitella [24]
3 years ago
11

Which material would you most likely to find in the c horizon

Physics
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

the answer would be parent material

Explanation:

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What will most likely happen if a light wave moves through a solid?
Kazeer [188]
In will most likely decrease its speed.
hope this helps.
8 0
3 years ago
Read 2 more answers
What is the net force on a car with a mass of 1000 kg if its<br> acceleration is 35 m/s^2?
VashaNatasha [74]

Answer:

3000N

Explanation:

divided to get answer

the force needed to accelerate the 1000kg car by 3m/s2 is 3000N

7 0
2 years ago
A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0
2 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
3 years ago
A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4
irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

mvL = (I + mL^2)\omega

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega

1.75 = [4.5 + 0.245]\omega

1.75 = 4.745\omega

\omega = 0.37 rad/s

so the final angular speed will be 0.37 rad/s

4 0
3 years ago
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