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Ann [662]
2 years ago
14

Quinn is testing the motion of two projectiles x and y by shooting them from a sling shot. What can we say best describes the mo

tion of the projectiles ? Assume air resistance is not a factor
Physics
2 answers:
Semmy [17]2 years ago
6 0

B. The vertical velocity of projectile Y is changing, and the horizontal velocity of projectile X is constant.

Studentka2010 [4]2 years ago
3 0

Explanation:

            A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path  subjected to gravity force only.

           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

             In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

You might be interested in
To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.
lozanna [386]

Answer:

a)

Explanation:

  • A block sliding down an inclined plane, is subject to two external forces along the slide.
  • One is the component of gravity (the weight) parallel to the incline.
  • If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

        F_{gp} = m*g* sin \theta (1)

       (taking as positive the direction of the movement of the block)

  • The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
  • When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

       F_{f} = \mu_{k} * N  (2)

  • The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

       N = m*g* cos \theta (3)

  • In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

        m*g* sin \theta =  \mu_{k} * m*g* cos \theta

  • As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
  • In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

        \mu_{k}  = tg \theta

8 0
2 years ago
Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
2 years ago
A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2
Ostrovityanka [42]
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
 We have then
 d ^ 2 = 16 ^ 2 + 12 ^ 2

 From here, we clear the value of d.
 We have then:
 d =  \sqrt{16 ^ 2 + 12 ^ 2} 

 Rewriting:
 d = \sqrt{256 + 144}
 d = \sqrt{400}
 d = 20 miles

 Answer:
 
The magnitude of the car's displacement is:
 
d = 20 miles
7 0
2 years ago
Read 2 more answers
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
2 years ago
Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
2 years ago
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