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lions [1.4K]
3 years ago
14

I dont understand pojectile motion can someone help with this

Physics
1 answer:
katrin2010 [14]3 years ago
3 0

When thrown from the balloon, the sandbag has a constant horizontal/forward velocity. As the text puts it, "the sandbag continues forward as if it had not been dropped." What this means is that the sandbag has no acceleration in the horizontal plane of motion, so it moves in this direction at a constant speed. So for any time t, the horizontal velocity is given by

v_x(t) = 12\dfrac{\rm m}{\rm s}

In the vertical plane of motion, the sandbag has no vertical velocity until it's released and allowed to fall. As it's falling, however, the sandbag is being pulled down by gravity and is accelerating with magnitude g=9.8\frac{\rm m}{\mathrm s^2}, so its vertical/downward velocity at time t is given by

v_y(t) = -gt

So, after 3 seconds have passed, the sandbag has horizontal velocity v_x = 12\frac{\rm m}{\rm s} and vertical velocity v_y = -g(3\,\mathrm s) = -29.4\frac{\rm m}{\rm s}. If the question is specifically asking about speed you would just ignore the negative signs.

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Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par
Murrr4er [49]

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0
3 years ago
Write the formula used to calculate the weight of an object
Leno4ka [110]

One possible formula is

<em>Weight = (mass of the object) x (acceleration of gravity where the object is).</em>


5 0
3 years ago
Read 2 more answers
A bucket of water with mass 5 kg sits on the ground with a coefficient of static friction of 0.35. What is the maximum force of
allochka39001 [22]

Answer:

The force of static friction is 17.15 N

Explanation:

It is given that,

Mass of the bucket, m = 5 kg

The coefficient of static friction is, \mu=0.35

We need to find the maximum force of static friction. It is given by :

F=\mu mg

F=0.35\times 5\ kg\times 9.8\ m/s^2

F = 17.15 N

So, the force of static friction is 17.15 N. Hence, this is the required solution.

6 0
4 years ago
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
4 years ago
A body of mass 50kg is placed on a wall of height 1.5m what is the potential energy​
DiKsa [7]

Answer:

PE=mgh

=50×1.5×9.8

=735

Explanation:

need thanks and make me brainiest if it helps you

8 0
3 years ago
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