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3 years ago

A sprinter leaves the starting blocks with an acceleration of 4.5m/s^2. What is the sprinter's speed 2 s later

Physics

2 answers:

mylen [45]3 years ago

6 0

Answer:

9 m/s

Explanation:

For a uniformly accelerated motion, the speed at time t is given by:

$v(t) = u +at$

where

u is the initial speed

a is the acceleration

t is the time

In this problem, we have:

u = 0 m/s is the initial speed

a = 4.5 m/s^2 is the acceleration

t = 2 s is the time

Substituting into the equation, we find

$v(2 s)=u+at=0 +(4.5 m/s^2)(2 s)=9 m/s$

tangare [24]3 years ago

4 0

V=v0+at
v=0 m/s + 4.5 m/s^2 * 2 s
v=9 m/s

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If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

2 years ago

Bob is pulling a 30 kg filing cabinet with a force of 200 N , but the filing cabinet refuses to move. The coefficient of static

Arada [10]

Answer:

<em>Magnitude of the Frictional force is 200 N</em>

Explanation:

The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.

Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.

Frictional force = Force applied

Force applied = 200 N

Therefore, the magnitude of the friction force on the filing cabinet is 200 N

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3 years ago

1hp is written in electric motor what does it mean

Angelina_Jolie [31]

Explanation:

The horsepower (hp) is a unit in the foot-pound-second ( fps ) or English system, sometimes used to express the rate at which mechanical energy is expended. It was originally defined as 550 foot-pounds per second (ft-lb/s). A power level of 1 hp is approximately equivalent to 746 watt s (W) or 0.746 kilowatt s (kW).

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2 years ago

Why would an orange roll of your cafeteria tray if you stopped suddenly

Vikentia [17]

Because the frictional force between the orange skin peel is great enough when you are walking for it to be carried on the tray, along with the gravitational force downwards onto the tray. When you stop, the force that you exerted moving forward it the same as on the tray and on the orange. So when you stop, the force is still on the orange as the same velocity as your we’re traveling, while the tray and you stop.

6 0

3 years ago