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bearhunter [10]
2 years ago
11

4. A 1300 kg car makes a turn with a 45 m radius. Coefficient of static

Physics
1 answer:
asambeis [7]2 years ago
3 0

Answer: v = 11 m/s

Explanation:

If we ASSUME that the road is horizontal, not banked

The maximum friction force available is μmg

This will need to supply all of the required centripetal force

mv²/R = μmg

v = √(μgR) = √(0.28(9.8)(45)) = 11.1121...≈ 11 m/s

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A 0.0780 kg lemming runs off a
kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
2 years ago
Please help me thank you !!!!
Fantom [35]

Answer:

<em>UP</em>

Explanation:

heat flows from higher level to lower level

( higher concentration to lower concentration )

and since temperature in above block is less than the lower block, the heat will flow from lower block to higher block .

( Up )

5 0
3 years ago
Which planets are visible without a telescope
Annette [7]

The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.

3 0
3 years ago
The diagram below shows a periodic wave. Which two points on the wave are 180 degrees out of phase?
Ksju [112]

Answer:

A and C are 180 deg out of phase (opposite points on a 360 deg wave)

3 0
3 years ago
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel
Salsk061 [2.6K]

Answer:

If the particle is an electron E_y = 3.311 * 10^3 N/C

If the particle is a proton, E_y = 6.08 * 10^6 N/C

Explanation:

Initial speed at the origin, u = 3 * 10^6 m/s

\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, m_e = 9.1 * 10^{-31} kg

Mass of a proton, m_p = 1.67 * 10^{-27} kg

The electric field intensity along the positive y axis E_y, can be given by the formula:

E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\

If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C

If the particle is a proton:

E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 6.08 * 10^6 N/C

8 0
3 years ago
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