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olga nikolaevna [1]
2 years ago
14

How can we take advantage of waste materials​

Physics
1 answer:
Vinvika [58]2 years ago
6 0

Answer:

we can incinerate waste materials and use the product as fuel rather than dumping them. It may produce carbon dioxide but we can create a machine that can convert carbon into energy.

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Are amplitude and wavelength related?<br> A: Yes<br> B: No
Trava [24]

The answer is A.Yes

Explanation:

The amplitude of a wave is the height of a wave as measured from the highest point of the wave to the lowest on the wave.

4 0
2 years ago
Scientific theories are deductive in nature.?
dolphi86 [110]

Answer:

deductive reasoning usually follows steps .

  • That is, how we predict what the observations should be if the theory were correct

8 0
2 years ago
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Kisachek [45]

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

3 0
3 years ago
(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,
Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = 2.677m/sec^2

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration a=1.35m/sec^2

Final speed v = 80 km/hr

80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec

From first equation of motion v =u+at

So t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration a=1.65m/sec^2

So t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2

As acceleration is negative so it is a deceleration

7 0
3 years ago
A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
son4ous [18]

Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

T = 1.83 h

now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{6.99}{2.89}

\theta = 67.5 degreeNorth of East

6 0
3 years ago
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