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3 years ago

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A skater increases her speed uniformly from 2.0 meters per second to 7.0 meters per second over a distance of 12 meters. the mag

nitude of her acceleration as she travels this 12 meters is

1 answer:

AVprozaik [17]3 years ago

6 0

<span>The magnitude of her acceleration as she travels this 12 meters is 1.875m/s^2</span>

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A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of the

taurus [48]

Answer:

The mechanical energy of the helicopter is $1.12\times 10^7\ J$ .

Explanation:

It is given that,

Mass of the helicopter, m = 3250 kg

Speed of the helicopter, v = 56.9 m/s

Position of the helicopter, h = 185 m

The energy possessed by an object due to its motion is called its kinetic energy. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3250\times (56.9)^2$

$E=5.26\times 10^6\ J$

The energy possessed by an object due to its position is called its potential energy. It is given by :

$E=mgh$

$E=3250\times 9.8\times 185$

$E=5.89\times 10^6\ J$

The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :

$M=5.26\times 10^6+5.89\times 10^6$

$M=11.15\times 10^6\ J$

or

$M=1.12\times 10^7\ J$

So, the mechanical energy of the helicopter is $1.12\times 10^7\ J$ . Hence, this is the required solution.

6 0

3 years ago

A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is its acceleration?

saul85 [17]

Alright let's start this off with our basic equation!

Accerlation (a) = ?

Initial velocity (V1)= 26 m/s
Final velocity (V2) = 0 ft/s because the car comes to a complete stop!

Time (t) = 6 seconds

The equation for acceleration is below!
a = $\frac{Final velocity - Initial Velocity}{time}$

So now, just plug in the values!
a = $\frac{0 - 26}{6}$
a = $\frac{-26}{6}$
a = -4.33 m/s²

Therefore, your acceleration is -4.33 m/s²!! Hope this helped and was one of the branliest answers :')

5 0

3 years ago

Isaac Newton’s investigations of gravity explained which truth?

Darina [25.2K]

C. Gravity acts on all objects in the universe!

8 0

2 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

2 years ago

3. What is the acceleration of a 10 kg mass pushed by a 5 N force?

insens350 [35]

Answer:

F=ma

Plug it in:

5=10a

5/10=(10a)/10

.5m/s²=a

Explanation:

Brainliest?

5 0

3 years ago

Read 2 more answers