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3 years ago

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A car starts from rest and accelerates at a constant rate after the car has gone 50 m it has a speed of 21 m/s what is the accel

eration of the car

1 answer:

Leya [2.2K]3 years ago

3 0

Answer:

4.41 m/s^2

Explanation:

(v_f)^2 - (v_i)^2 = 2a * change in distance

(21)^2 - (0)^2 = 2a * 50

a = (21^2)/(2*50)

a = 4.41 m/s^2

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The amount of gas that a helicopter uses is directly proportional to the number of hours spent flying. the helicopter flies for

igomit [66]

Answer:

The helicopter uses 35 gallons to fly for 5 hours.

Explanation:

The amount of gas that a helicopter uses for flying varies directly proportional to the number of hours spent flying.

g ∝ T

where g represents amount of gas and T time of flight.

Then,

$\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}$

The helicopter files 4 hours and uses 28 gallons of fuel.

Here, g₁= 28 gallons, T₁=4 hours

g₂=?, T₂=5 hours.

$\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}$

$\Rightarrow \frac{28}{g_2}=\frac{4}{5}$

⇒28×5= g₂×4

⇒ g₂×4=28×5

$\Rightarrow g_2=\frac{28\times 5}{4}$

$\Rightarrow g_2=35$ gallons

The helicopter uses 35 gallons to fly for 5 hours.

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3 years ago

What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = $9.8 m/s^2$

Mass of the object = m

Force = mass × acceleration

$510 N=m\times 9.8 m/s^2$

$m=52.04 kg\approx 52 kg$

The mass of an object is 52 kg.

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3 years ago

Read 2 more answers

You indicate that a symbol

Goryan [66]

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

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3 years ago

How does a mirror affect the path of light?

yulyashka [42]

Light rays change direction when they hit a mirror. The phenomenon is known as reflection. Light rays travels in a straight light. They strike the surface of the mirror at a particular angle called incident angle. It is the angle between the ray and normal at the point of contact. The rays leaves the surface making the same angle with the normal called reflection angle but in different direction.

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3 years ago

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

Now we can solve for $v$ . We find that

$v=20.8\,\dfrac{\mathrm m}{\mathrm s}$

In the case that the ratio of accelerations is actually

$\dfrac{a_2}{a_1}=1.66$

we would instead have

$\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)$

in which case we would get a velocity of

$v=24.4\,\dfrac{\mathrm m}{\mathrm s}$

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3 years ago