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Firdavs [7]
4 years ago
14

How is a quarter able to role on its edge?

Physics
1 answer:
Olegator [25]4 years ago
5 0
Idk its really weird because the ridges are making the quarter stand and therefore able to roll
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a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance
8_murik_8 [283]

The car's (average) acceleration would be

a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}

The car's position over time would be given by

x=v_0t+\dfrac12at^2

so that after 2.4 seconds, the car will have traveled a distance of

x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2

\implies x=77.5\,\mathrm m

7 0
3 years ago
Read 2 more answers
The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol
Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

8 0
3 years ago
Explain the types of root​
Alex_Xolod [135]

Answer:

there are two main type of root systems.

6 0
3 years ago
Read 2 more answers
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
BaLLatris [955]

Answer:

= 4.3 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of  the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)

=  8.459 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = (  9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³

= 4.3 × 10 ⁻¹⁴ m

3 0
4 years ago
The surface manifestation of a mantle plum is called a ________.
galina1969 [7]
L think is Asthenosphere
3 0
3 years ago
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