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maria [59]
2 years ago
5

A car is traveling 16 m/s East. If the car then speeds up at a constant acceleration, what is the direction of the car’s acceler

ation?
Physics
1 answer:
konstantin123 [22]2 years ago
8 0
The direction is east since its constant
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Does 3.60 x 10 ⁻² have 2 significant figures
Mandarinka [93]

Answer:

Number of Significant Figures: 2

The Significant Figures are 3 6

Explanation:

= 3.60 × 102

(scientific notation)

= 3.60e2

(scientific e notation)

= 360 × 100

(engineering notation)

(one)

= 360

(real number)

4 0
3 years ago
Before using wind energy to produce electricity, humans used wind energy to produce PLEASE ANSWER QUICKLY
topjm [15]

Answer:

Water

Explanation:

windmills were used to bring water to farms in ancient times.

8 0
3 years ago
A camper walks away from camp. He walks 23.72m north, then 39.25m south and
Arte-miy333 [17]

Answer:

89.53    North

Explanation:

Total N-S displacement

   23.72  -  39.25 + 105.06   =  + 89.53

5 0
1 year ago
Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is conne
lora16 [44]

Answer:

Energy stored in the capacitor will be 1.835\times 10^{-6}J

Explanation:

We have given edge length of capacitor = 1.2\times 10^2cm=1.2m

So area A = 1.2×1.2 = 1.44 m^2

Separation is given as d  1 mm = 10^{-3}m

We know that capacitance is given by C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F

Voltage is given as V = 12 volt

We know that energy stored is given by E=\frac{1}{2}CV^2=\frac{1}{2}\times 12.744\times 10^{-9}\times 12^2=1.835\times 10^{-6}J

3 0
3 years ago
Read 2 more answers
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