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Nat2105 [25]
4 years ago
9

A greyhounds velocity changes from 10 meters per second to 15 meters per second in 0. 5 second. What is the greyhounds average a

cceleration?
5 m/s^2
10 m/s^2
10 m/s
15 m/s
Physics
1 answer:
Radda [10]4 years ago
7 0
The answer is 10m/s^2
This is because a={v-u}/t 
a={15-10}/0.5=5/0.5=10m/s^2
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GREYUIT [131]
Many People<span> Are Averse to Management "</span>Control<span>" ... and then subsequent decisions about what to </span>do<span> is the essence of management coordination</span>
5 0
3 years ago
A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductiv
KonstantinChe [14]

Answer:

 P = 5280 W

Explanation:

The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is

    P = Q / t = k A (T_{h} -T_{c}) / L

Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of ​​the window, T the temperature and L is glass thickness

Let's calculate the window area

    A = l * a

    A = 2.0 1.0

    A = 2.0 m²

Let's replace

    L = 0.5 cm (1 m / 100 cm) = 0.005 m

    P = 0.8 2 (20.5 - 4) / 0.005

    P = 5280 W

7 0
3 years ago
A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b
guapka [62]
<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=200m/s is the bullet's initial speed

\theta=0 because we are told the bullet is shot horizontally

t is the time since the bullet is shot until it hits the ground

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=1.5m  is the initial height of the bullet

y=0  is the final height of the bullet (when it finally hits the ground)

g=9.8m/s^{2}  is the acceleration due gravity

<h2>Part (a):</h2>

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}   (3)

0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}   (4)

Finding t:

t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}   (5)

Then we have the time elapsed before the bullet hits the ground:

t=0.553s   (6)

<h2>Part (b):</h2>

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

x=V_{o}cos\theta t   (1)

Substituting the knonw values and the value of t found in (6):

x=200m/s.cos(0)(0.553s)   (7)

x=200m/s(0.553s)   (8)

Finally:

x=110.656m  

4 0
3 years ago
Need help with this! <br><br> just scienceeeeeee
LiRa [457]

Answer:

7)  λ = 0.5 m,  8)  f = 4.8 10¹⁴ Hz

Explanation:

The speed of an electromagnetic wave is

          c = λ f

where c is the speed of light in vacuum c = 3 10⁸ m / s

7) indicate the frequency f = 6.0 10⁸ Hz

we do not know the wavelength

         λ = c / f

       

we calculate

        λ = 3 10⁸ / 6.0 10⁸

        λ = 0.5 m

8) indicate the wavelength  λ = 6.25 10-7 m

we do not know the frequency

         f = c / λ

we calculate

        f = 3 10⁸ / 6.25 10⁻⁷

        f = 0.48 10¹⁵ Hz

        f = 4.8 10¹⁴ Hz

8 0
3 years ago
A___ is a large,bright,gaseous feature extending outward from the sun’s surface,often in a loop shape A)Sunspots B)Prominence C)
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B

A prominence is a large, bright, gaseous feature extending outward from the Sun's surface, often in a loop shape. Prominences are anchored to the Sun's surface in the photosphere, and extend outwards into the Sun's corona.

7 0
3 years ago
Read 2 more answers
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