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3 years ago

if a student lifts their weight of 450 newtons up a set of stairs 5 meters high, how much work did they do?

Physics

2 answers:

Gnoma [55]3 years ago

8 0

Answer:

work done = 2250 j

Explanation:

work done = force × displacement

=》

$w.d. = 450 \times 5$

=》work done = 2250 joules

irina1246 [14]3 years ago

4 0

Answer:

2,250J

Explanation:

W = Fs = (450)(5) = 2,250

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The graph represents velocity over time. What is the acceleration? –0.4 m/s2 –0.2 m/s2 0.2 m/s2 0.4 m/s2

Olenka [21]

Missing graph. I attach it in the answer.

In a uniformly accelerated motion, the velocity at time t is given by:
$v(t)=at$
where a is the acceleration and t is the time.

Given the previous equation, if we plot v(t) versus t, we find a straight line; moreover, a (the acceleration) represents the slope of the curve.

Looking at the graph, we see that when the time goes from 10 s to 20 s, the velocity increases from 4 m/s to 6 m/s. Therefore the slope of the curve is
$a= \frac{\Delta v}{\Delta t}= \frac{6 m/s-4 m/s}{20 s-10 s}= \frac{2 m/s}{10 s}=0.2 m/s^2$
and this corresponds to the acceleration.

So, the correct answer is <span>0.2 m/s2.</span>

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4 years ago

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A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?

kompoz [17]

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

7 0

3 years ago

Why is cathode positive in leclanche cell??

laila [671]

Answer:

Because the zinc is reluctant

Explanation:

A leclanche cell contains a conducting solution (electrolyte) of ammonium chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide (oxidizer), and an anode (negative terminal) of zinc (reductant).

As the Zn2+ ions move away from the anode, leaving their electrons on its surface,

Zn → Zn2+ + 2e−

the anode becomes more negatively charged than the cathode. When the cell is connected to an external electrical circuit, the excess electrons on the zinc anode flow through the circuit to the carbon rod, the movement of electrons forming an electric current.

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

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3 years ago

It took 500 newtons of force to push a car 4 meters. How much work was done?

Ede4ka [16]

Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J

3 0

2 years ago