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shepuryov [24]
2 years ago
12

A car drives toward the right over the top of a hill, as shown below. An illustration of car at the top of a hill pointing right

and looking at the side of the car. The is a vector up labeled Z, a vector down labeled X, a vector left labeled Y and a vector right labeled W. Which arrow represents the centripetal acceleration? W X Y Z

Physics
2 answers:
Sav [38]2 years ago
6 0

Answer: X

Explanation:

This situation can be illustrated as a car in circular motion (image attached).

In circular motion the acceleration vector \vec{a} is always directed toward the center of the circumference (that's why it's called centripetal acceleration).

So, in this case the arrow labeled X is the only that points toward the center, hence it represents the car's centripetal acceleration

kherson [118]2 years ago
3 0

Answer:

X

Explanation:

Because Centripetal Force pulls it towards the circle's center

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A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it
Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

  • 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

  • v = v₀ + at

Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
  • t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

  • 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
  • Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0
1 year ago
A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t
LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

4 0
2 years ago
Honeybees can see light in the ________ range of the electromagnetic spectrum.
konstantin123 [22]
Humans can see wavelengths in the visible part of the electromagnetic spectrum. That is the range of approximately 400 - 700 nm. Honeybees can see visible light and about 100 nm more in the ultraviolet part of the electromagnetic spectrum. That is approximately 300 - 700 nm. 
4 0
2 years ago
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle
liq [111]

Answer:

0.0031792338 rad/s

Explanation:

\theta = Angle of elevation

y = Height of balloon

Using trigonometry

tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta

Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

6 0
2 years ago
A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical loops of thread A and B placed at 20cm and 30cm f
podryga [215]

Answer:

(a) 29 cm

(b) 43.5 cm

Explanation:

(a) when loop A is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 70 cm mark

-2 N at x

Taking the sum of the torques about B:

∑τ = Iα

(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0

18 Ncm − 2 N (x − 70 cm) = 0

2 N (x − 70 cm) = 18 Ncm

x − 70 cm = 9 cm

x = 79 cm

The distance from the center is |50 cm − 79 cm| = 29 cm.

(b) when loop B is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 20 cm mark

-2 N at x

Taking the sum of the torques about A:

∑τ = Iα

(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0

-27 Ncm − 2 N (x − 20 cm) = 0

2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

The distance from the center is |50 cm − 6.5 cm| = 43.5 cm

3 0
2 years ago
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