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Wittaler [7]
3 years ago
7

The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?

Physics
1 answer:
Talja [164]3 years ago
7 0

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

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Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.
irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0
3 years ago
If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?
Otrada [13]

Answer: 0.2 hour

Explanation:

Speed = Distance / Time

Time = Distance / Speed

= 12/60

= 0.2 hour

8 0
3 years ago
Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t
lara [203]

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

     A = 4π R₀²

     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0
3 years ago
A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
What will most likely happen if a sound wave moves from the air through a solid?
stich3 [128]

Answer:

It will increase in speed

Explanation:

6 0
3 years ago
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