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ioda
3 years ago
11

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ration g is directed 1 Upward and ts value changes with height 2 Upward and its value changes with height 3 There is no accelaration 4 Downward and its value is constant 5 Downward and ts value changes with height
Physics
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for y, and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

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Please help me. I don’t understand this.
guajiro [1.7K]

Answer:

Ft = 0[N]

Explanation:

To understand this problem we must perform an analysis of forces on the X axis, which coincides with the axis of forces of dogs.

In this way performing a sum of forces on the X-axis we will have (newton's third law):

F_{rigth}-F_{left}+F_{total} =0\\\  200-200+F_{total} =0\\F_{total}=0

From this analysis we can see that the resulting or total force is equal to zero, since there is no movement.

8 0
4 years ago
A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
Sati [7]

(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

6 0
3 years ago
Can someone please help me and write the answers? i need this and it’s really important!!
eduard

-- Kinetic energy is the energy of mass in motion.  The amount is determined by the mass of whatever is moving, and its speed.

-- Potential energy is the energy that's stored up in some form, not being used yet but ready to be used when you want it.

For example, one form of it is <u><em>chemical</em></u><em> </em>potential energy, like in a battery, or a match.  You get the energy out of a battery when you connect it to a motor or a light.  You get the energy out of as match when you make the tip hot and it flares up.

This question is asking about <u><em>gravitational</em></u> potential energy.  An object has stored energy just by being up high, like a bowling ball on a shelf.  You get the energy out of it just by dropping it ... possibly enough to crack the floor !

The amount of this kind of potential energy is determined by the mass of the object, and how high up it is.

-- Getting the answers from other people doesn't help you a bit, until you understand them and can answer the question on your own.

5 0
3 years ago
If the maximum length is 0.3 meters from the equillibrium point what is the and the spring oscillates 100 per minute what is it'
kaheart [24]
20.......................................
5 0
4 years ago
the net force on a vehicle is accelerating at a rate of 1.5 milliseconds is 1800 Newtons. What is the mass of the vehicles neare
Goryan [66]

Answer:

The mass is 1200 kilograms

Explanation:

Because Force is equal to mass times acceleration (F=m×a)

F=m×a

1800N=?×1.5

1800÷1.5=1200

1800N=1200Kg×1.5

7 0
4 years ago
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