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vichka [17]
4 years ago
9

Suppose that you are headed toward a plateau 4040 m high. If the angle of elevation to the top of the plateau is 2020degrees°​,

how far are you from the base of the​ plateau?

Physics
1 answer:
Lelu [443]4 years ago
3 0

Answer :

The distance is 109.89 m.

Explanation :

Given that,

Height = 40 m

Angle = 20°

We need to calculate the base of the​ plateau

Using formula of angle

\tan\thata=\dfrac{h}{x}

Where, h = height

x = base

\theta = angle

Put the value into the formula

\tan20^{\circ}=\dfrac{40}{x}

x=\dfrac{40}{tan20^{\circ}}

x =109.89\ m

Hence, The distance is 109.89 m.

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3 0
3 years ago
3. A rocket is launched at an angle of 53 degrees above the 1 point
irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

3 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
A swimmer pushing off from the wall of a pool exerts a force of 1 newton on the wall. What is the reaction force of the wall on
8090 [49]

Answer: 1 Newton

Explanation:

"Every action has an equal and opposite reaction."

Please mark as Brainliest if it is correct.

3 0
3 years ago
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