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MAVERICK [17]
2 years ago
14

A car speeds up from 14 meters per second to 21 meters per second in 6 seconds. Whats the acceleration and the distance passed w

ithin that time?
Physics
1 answer:
Solnce55 [7]2 years ago
7 0

Answer:

a = 1.666... m/s²

Explanation:

a = v2 - v1 / t2 - t1

a = 21m/s - 14m/s / 6s - 0s

a = 7m/s / 6s

a = 1.666... m/s²

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20 points+ Brainliest
Helen [10]

Answer:

The answer is B. hope this helps

Explanation:

3 0
3 years ago
After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
The elements carbon, nitrogen, and oxygen are'll part of the same _____ on the periodic table?
andreev551 [17]
C, N and O all belong to the same period, in which it's 2nd Period.
4 0
3 years ago
Read 2 more answers
The speed of light is 3.0 x 108 m/s. How far does light travel in one day? Remember that you should first convert the day into s
Anna35 [415]

Answer:

2.59\times 10^{13}\ m

Explanation:

The speed of light is 3.0 x 10⁸ m/s

We need to find the distance traveled by it in one day.

1 day = 24 hours

1 day = 24 × 3600 seconds

= 86400 s

Now, using the formula of speed to find the distance covered as follows :

d=v\times t

Putting all the values, we get :

d=3\times 10^8\ m/s\times 86400\ s\\\\d=2.59\times 10^{13}\ m

Hence, light will cover 2.59\times 10^{13}\ m of distance in one day.

6 0
3 years ago
A thin layer of oil of refractive index 1.22 is spread on the surface of water (n = 1.33), If the thickness of the oil is 275 nm
ANTONII [103]

Answer:

6.71×10⁻⁷ m

Explanation:

Using thin film constructive interference formula as:

<u>2×n×t = m×λ</u>

Where,

n is the refractive index of the refracted surface

t is the thickness of the surface

λ is the wavelength

If m =1

Then,

2×n×t = λ

Given that refractive index pf the oil is 1.22

Thickness of the oil = 275 nm

Also, 1 nm = 10⁻⁹ m

Thickness = 275×10⁻⁹ m

So,

Wavelength is :

<u>λ= 2×n×t = 2× 1.22 × 275×10⁻⁹ m = 6.71×10⁻⁷ m</u>

8 0
3 years ago
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