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Alexxx [7]
3 years ago
9

A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the

air and lands a horizontal distance of 32m from where it was released. If the stone returns to the same height from which it was thrown, which of the following claims best describes the motion of the air through the stone's trajectory? There must be a horizontal wind in the direction of the stone's motion, because ignoring air resistance when alculating the horizontal range would yield a value less than 32m There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value less than 32m. (C) There is no horizontal wind, because ignoring air resistance when calculating the horizontal range would yield a value equal to 32m. There must be a horizontal wind in the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32m. There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32m
Physics
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

Option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile = 20 m/s

θ = angle above the horizontal at which the projectile was launched = 30°

g = acceleration due to gravity = 9.8 m/s²

R = (30² sin 60°) ÷ 9.8

R = 78.53 m

So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Hope this Helps!!!

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VARVARA [1.3K]

Answer:

the correct one is D,  

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

          E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

          c = λ f

         

we substitute

          E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation                E = 10⁻² eV

visible radiation                   E = 1 to 3 eV

radiation Uv                         E = 3 to 6 eV

X-ray                                    E = 10 eV

 

gamma rays                         E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

6 0
2 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
What is a fault?
Free_Kalibri [48]
C) A crack in earth's crust where movement occurs. An example of this is the San Andreas Fault.

A) is repulsion
B) continental crust (lighter crust)
D) Hotspot
3 0
3 years ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4
Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 m/s^{2}, A moves downwards whereas B moves upwards.

\alpha=6.125 rad/s^{2}

Tension on side A = 4.5 × g= 44.1 m/s^{2}

Tension on side B= 2.0 × g=  19.6 m/s^{2}

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be T_{2} and the tension attached with Block B be T_{1} .

Tensions will be only be due to the weight of the blocks as no other force is present.

T_{2} = 4.5 × g= 44.1 m/s^{2}

T_{1} = 2.0 × g=  19.6 m/s^{2}

Now, lets make a torque equation about the center of the wheel and find the alpha

T_{2}×R- T_{1}×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m  and

           Alpha(\alpha)= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/m^{2}

(44.1-19.6)R=0.400\alpha

\alpha = \frac{24.5 * 0.100}{0.400}

\alpha=6.125 rad/s^{2}

Acceleration = R ×\alpha

                    = 0.1 * 6.125

                    =0.6125 m/s^{2}

Accelerations of both the sides is 0.6125 m/s^{2}, A moves downwards whereas B moves upwards.

7 0
3 years ago
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