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uysha [10]
3 years ago
10

• ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.

Physics
1 answer:
nignag [31]3 years ago
5 0

Answer:

45.6 m at 80.5^{\circ} south of west

Explanation:

Let's take the north-south direction as y-direction (with south being positive) and east-west direction as x-direction (with west being positive). Therefore, the two components of Cody's motion are:

- d_y = 45.0 m (south)

- d_x = 7.50 m (west)

Since they are perpendicular, the magnitude of the net displacement can be calculated by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{7.50^2+45.0^2}=45.6 m

The direction instead can be measured as follows:

\theta = tan^{-1} (\frac{d_y}{d_x})=tan^{-1}(\frac{45.0}{7.50})=80.5^{\circ}

And given the convention we have used, this angle is measured as south of west.

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Calculate the electric field on the surface of a charged sphere with a radius of 5cm and charge 10μC. Thanks
victus00 [196]

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is σ=Q/4πR2, then indeed the field at the surface is σ/ϵ0 as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.

Explanation:

4 0
2 years ago
A 1,000 kg ball traveling at 5 m/s would have
jonny [76]

Answer:

15 because 5×5×5 is the same thing as 5×3 which equals to 15

6 0
3 years ago
Read 2 more answers
A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
3 years ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
iren [92.7K]

Answer:

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

3 0
3 years ago
Sound travels about 750 miles per hour. if you stand in a canyon and sound a horn, you will hear an echo. suppose it takes about
diamong [38]
The answer to this question would be:3850ft

To answer this question, you need to convert the speed velocity from miles/hour into feet/second. The equation would be: 750 miles/hour x 5280 foot/mile x 1 hour/3600second = 1100 ft/s
Then multiply the time with the velocity= 3.5 second x 1100 ft/s= 3850ft
5 0
4 years ago
Read 2 more answers
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