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antoniya [11.8K]
3 years ago
14

Convert 4 radians to degrees A. 242.6° B. 9.6° C. 229.2° D. 6.8°

Physics
2 answers:
muminat3 years ago
8 0
The central angle of a circle is 360° or 2π radians.

Therefore
1 radian = (360 degrees)/(2π radians) = 180/π degrees/radian.
4 radians = (4 radians)*(180/π degrees/radian) = 229.18 degrees.

Answer: C.  229.2°
saveliy_v [14]3 years ago
5 0
C 229.2 is the answer because of pie
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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
3 years ago
To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x
strojnjashka [21]

Answer:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

5 0
3 years ago
To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?
Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0
3 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0
3 years ago
Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0
3 years ago
Read 2 more answers
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