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nadezda [96]
3 years ago
15

¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?

Physics
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

s=0+\frac{1}{2}(5)(10)^2=250 m

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A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave
Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0
3 years ago
Which vector is the sum of vectors a and b
jeka57 [31]

Answer:

C

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Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi
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Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

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Applying ohm's law,

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Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

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dlinn [17]

Answer:

c

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Step-by-step explanation:

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