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BlackZzzverrR [31]

3 years ago

7

Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component

is unchanged, what is the effect on the time of flight

1 answer:

Katarina [22]3 years ago

3 0

Answer:

Explanation:

Time of flight = 2 x u sinα / g where u sinα is vertical component of projectile's velocity u .

So Time of flight = 2 x vertical component / g

vertical component = constant

g is also constant so

Time of flight will also be constant .

It will remain unchanged .

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SVEN [57.7K]

When you add up (20N down) and (10N up),
you get a sum of (10N down).

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6 0

4 years ago

A 120N box is being pushed across the surface that has a frictional force of 3N. If the box is being pushed with a constant forc

VashaNatasha [74]

Answer:

Net Force = 10 N

acceleration: $a=0.817 \frac{m}{s^2}$

Explanation:

1) The net force on the box is the applied force (F = 13 N) minus the friction force (f = 3 N), Therefore, net force = 10 N

See attached free body diagram.

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$weight=m*g\\120 N = m*9.8\frac{m}{s^2} \\\frac{120}{9.8} kg = m\\m=12.24 kg$

Now we can find the box's acceleration by using the equation for the net force:

$F=m*a\\10N = 12.24 kg * a\\\frac{10}{12.24} \frac{m}{s^2} =a\\a=0.817 \frac{m}{s^2}$

5 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

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Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

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$F_x=kx$

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$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

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5 0

3 years ago

If a motorcyclist accelerates from rest to 40 m/s in 8 seconds, what is their acceleration?

IgorC [24]

Answer:

5 m/s²

Explanation:

Given:

v₀ = 0 m/s

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t = 8 s

Find: a

a = (v − v₀) / t

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a = 5 m/s²

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We're millions of miles away from mars

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