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WARRIOR [948]
3 years ago
15

What is the acceleration of a car that is going at a steady speed of 60 mph?

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

0

Explanation: you are only accelerating if you are slowing down, speeding up, or changing direction

You might be interested in
Estimate the boiling point of water at a pressure of 1.1 atmosphere
Nezavi [6.7K]
It is about 100oC at a pressure of 1.1 atmosphere. Hope this helps.
4 0
3 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
2 years ago
What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s
Ilia_Sergeevich [38]

Speed of wave is given as

v = 5 m/s

Wavelength of the wave is given as

\lambda = 20 m

now from the formula of wave time period we can say

speed = \frac{wavelength}{time period}

5 = \frac{20}{T}

T = \frac{20}{5}

T = 4 s

so it will have time period of T = 4 s

7 0
3 years ago
On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai
xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,

the horizontal component of this initial velocity is given by

v_x = v\cos \theta

v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}

The vertical component is given by

v_y = v\sin\theta

v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}

3 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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