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weqwewe [10]
2 years ago
14

In which situation is the acceleration of the car negative?

Physics
2 answers:
Schach [20]2 years ago
6 0
B) The velocity of a car reduced from 50 km/h to 35 km/h over one minute
olga_2 [115]2 years ago
5 0
The most appropriate answer is B !!

as the velocity of car is decreased in one min, from 50 to 35 , so there must be deacceleration !! or say negative acceleration !!

so your answer is B !!
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When is a zero not significant?
maria [59]

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0
3 years ago
Read 2 more answers
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
Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa
-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              \texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C

Similarly for capacitor 2

               \texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C

Capacitance of the second capacitor = 2C

6 0
3 years ago
In order to increase the amount of work done, we need to:
Nadusha1986 [10]
Option (D) is the correct one.


In order to increase the amount of work done, we need to increase the force applied to the object.
8 0
3 years ago
What causes glowsticks to give off light?
iren2701 [21]
Phenyl oxalate ester is responsible for the luminescence in aglow stick<span>. The reaction with hydrogen peroxide </span>causes<span> the liquid inside a </span><span>glow stick to glow</span>
3 0
3 years ago
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