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3 years ago

a toy airplane is flying at a speed of 8 m/s with an acceleration of 0.9 m/s^2. How fast is it flying after 2 seconds?

Physics

1 answer:

Marat540 [252]3 years ago

4 0

Answer:

it will be flying 1.8 m/s

Explanation:

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Which expression is equivalent to 5(2+7)? 0 2(5+7) O 2 + 7(5) O 5(2)+7 O 5(2)+5(7)

professor190 [17]

Answer:

5(2)+5(7)

Explanation:

5(2)+5(7)=10+35=45

6 0

3 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago

A cyclist is travelling eastwards at a velocity

VMariaS [17]

Answer:

0.02 m/s^2

Explanation:

change in velocity= 4.5m/s - 2.3m/s = 2.2 m/s

acceleration= change in velocity/change in time

acceleration= 2.2/120= 0.0183

= 0.02 (to 2 significant figures)

8 0

2 years ago

A flying squirrel has a mass of 0. 765 kg. He jumps off the tree and hits the ground with 125 joules of energy. Determine how hi

uranmaximum [27]

Answer:

Height, h = 16.67 m

Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

$E=mgh$

h is height up on the tree the squirrel was when it jumped

$h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m$

So, the squirrel will go to a height of 16.67 m.

6 0

3 years ago

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago