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Sergio039 [100]
3 years ago
5

A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of

time? A. Its acceleration is 1.25 meters/second. B. Its acceleration is 5 meters/second. C. Its acceleration is 0.8 meters/second. D. Its acceleration is 1.25 meters/second2. E. Its acceleration is 20 meters.
Physics
2 answers:
Natali5045456 [20]3 years ago
7 0

Answer:

The acceleration of the car is  a=1.25\ m/s^2.

Explanation:

It is given that,

Initial speed of the car, u = 0

Final speed of the car, v = 5 m/s

Time, t = 4 s

<em>We need to find the average acceleration over this period of time. It is given by the rate of change of velocity. It is given by :</em>

a=\dfrac{v-u}{t}

a=\dfrac{(5-0)\ m/s}{4\ s}

a=1.25\ m/s^2

So, the acceleration of the car is  a=1.25\ m/s^2. Hence, the correct option is (D).

disa [49]3 years ago
5 0

The average acceleration is

\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}

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you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
HELP
Ivanshal [37]

Answer:

The velocity is 60 km/hr.

Explanation:

<h3><u>Given:</u></h3>

Displacement (d) = 480 km = 48000 m

Time (t) = 8 Hours = 480 minute

Velocity (v) = ?

Now,

Velocity = Displacement ÷ Time

v = d/t

v = 480/8

v = 60 km/hr

Thus, The velocity is 60 km/hr.

<u>-TheUnknownScientist 72</u>

5 0
2 years ago
For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution
Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}

<u>a = 8.06 m/s²</u>

3 0
2 years ago
What is the apparent magnitude of the brightest star in the Big Dipper?
alex41 [277]
The magnitude of Alioth ( the brightest star in the big dipper ) is 1.76 and it is about 81 light years distant from Earth. 
4 0
3 years ago
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