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Yuki888 [10]
3 years ago
15

Two spheres of the same size and mass roll down an incline. Sphere A is hollow and Sphere B is solid with uniform density. Which

reaches the bottom of the incline first?
Make sure to EXPLAIN your answer!

(a) Sphere A
(b) Sphere B
(c) At the same time.
Physics
1 answer:
sweet [91]3 years ago
8 0
If the spheres are have to have the same and the same size, it is expected that they will have the same surface area. The only factor that affects the descent of the object through gravity on Earth is the surface area that is exposed to the resistance of air. For this case, it is expected that they will reach the bottom of the inclined plane together. The answer is C. 
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Does an increase in velocity necessarily mean an increase in acceleration?
Vinil7 [7]
We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then, 
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity

In short, Your Answer would be "Yes"

Hope this helps!
3 0
2 years ago
Read 2 more answers
The mass of a star is 1.61·1031 kg and its angular velocity is 1.60E-7 rad/s. Find its new angular velocity if the diameter sudd
Tcecarenko [31]

Answer:

ω₂ = 1.9025 x 10⁻⁶ rad/s

Explanation:

given,

mass of star = 1.61 x 10³¹ kg

angular velocity = 1.60 x 10⁻⁷ rad/s

diameter suddenly shrinks = 0.29 x present size

      r₂  = 0.29 r₁

using conservation of angular momentum

I₁ ω₁ = I₂ ω₂

(\dfrac{2}{5}mr_1^2)\omega_1=(\dfrac{2}{5}mr2^2)\omega_2

r_1^2\times \omega_1=r_2^2\times \omega_2

r_1^2\times 1.60\times 10^{-7}=(0.29r_1)^2\times \omega_2

1.60\times 10^{-7}=0.0841\times \omega_2

\omega_2=\dfrac{1.60\times 10^{-7}}{0.0841}

   ω₂ = 1.9025 x 10⁻⁶ rad/s

5 0
3 years ago
a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2
Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0
3 years ago
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between th
Sauron [17]

Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

<u>V = 0.45 Volts</u>

4 0
3 years ago
A ball is thrown vertically downward from the top of a 30.6-m-tall building. The ball passes the top of a window that is 10.7 m
Gekata [30.6K]

Answer:

v = 19.6 m/s

Explanation:

Height of building = 30.6 m.

Height of window from the ground level= 10.7 m.

Acceleration due to gravity = 9.8 \frac{m}{s^2}

At initial condition ball at rest condition so u= 0 m/s.

Lets take when passes through the window ,velocity is v.

Here acceleration is constant so we can apply motion equation .

We know that

v= u + a t

So by putting the values

v = 0 +9.8 x 2

v = 19.6 m/s

So the velocity of ball is 19.6 m/s when passes through the window after 2 s.

7 0
3 years ago
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