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Dmitry_Shevchenko [17]
3 years ago
13

A velocity vs. time graph is shown. A graph titled Velocity versus Time shows time in seconds on the x axis, numbered 0 to 5, ve

locity in meters per second on the y axis, numbered 0 to 20. A line starts at the origin and ends at (5, 20). What is the acceleration of the object?
Physics
2 answers:
masya89 [10]3 years ago
8 0

Answer:

a+4 m ...

Explanation:

Lisa [10]3 years ago
5 0

Answer:

a = 4 \frac{m}{s {}^{2} }

Explanation:

u for velocity

t for time

a for acceleration

u = a \times t \\ 20 = a \times 5 \\ a =  \frac{20}{5}  \\ a = 4

That's it! Hope I helped!

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6 0
3 years ago
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Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont
SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

T = \frac{2 u Sin\theta }{g}

T = \frac{2 \times 20 \times Sin35 }{9.8}

T = 2.34 second

(b) The formula for the maximum height is given by

H = \frac{u^{2} \times Sin^{2}\theta }{2g}

H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}

H  = 6.71 m

(c) The formula for the range is given by

R = \frac{u^{2} \times Sin 2\theta }{g}

R = \frac{20^{2} \times Sin 2 \times 35}{9.8}

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0
3 years ago
Match the term with the appropriate image
Tamiku [17]

First figure shows the object position

Second shows the image position

Third shows the focal length.

8 0
3 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
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Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
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a baseball pitcher throws a fastball at 42 meters per second. if the batter is 18 meters from the pitcher, approximately how muc
lapo4ka [179]
T=D/v = 18/42 = 43 seconds
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