Which is an example of projectile motion?

All objects, regardless of their mass, fall with the same rate of acceleration on

tamaranim1 [39]

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

4 0

3 years ago

Read 2 more answers

Please need help on this

ale4655 [162]

the answer is the third one down

7 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

Now the distance between the charge at at origin and the second charge:

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

Now the distance between the charge at at origin and the third charge:

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

Now the force due to second charge:

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

Now the force due to third charge:

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

Now the its horizontal component:

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

Now the its vertical component:

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Aleksandr [31]

"2 km/hr/s" means that in each second, its engines can increase its speed by 2 km/hr.

If it keeps doing that for 30 seconds, its speed has increased by 60 km/hr.

On top of the initial speed of 20 km/hr, that's 80 km/hr at the end of the 30 seconds.

This whole discussion is of speed, not velocity. Surely, in high school physics,
you've learned the difference by now. There's no information in the question that
says anything about the train's direction, and it was wrong to mention velocity in
the question. This whole thing could have been taking place on a curved section
of track. If that were the case, it would have taken a team of ace engineers, cranking
their Curtas, to describe what was happening to the velocity. Better to just stick with
speed.

4 0

4 years ago

A wire of density p is tapered so that its cross-sectional area varies with x according to

DochEvi [55]

The wave speed at the origin is v = 8.31 m/s

Given data

A=1.00× 10⁻⁵ x + 1.00× 10⁻⁶

A is in meters squared and x is in meters

tension in the wire is T

T = 24.0

x = 10.0m

density of aluminum = 2700

<h3>calculating for the speed of the wave at the origin</h3>

wave speed = v = sqrt ( T / mass per unit length)

mass per unit length = m / L

density = m / v

volume = v = A * L

density * A = m / L

wave speed = v = sqrt ( T / density * A )

v = sqrt ( 24 / ( 2700 * .00× 10⁻⁵ x + 1.00× 10⁻⁶ ) )

v = 8.31 m/s

Read more on speed of wave here: brainly.com/question/12969690

#SPJ4

4 0

2 years ago