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VashaNatasha [74]
2 years ago
14

Describe three different ways to change your velocity when youre riding a bike?

Physics
2 answers:
stepladder [879]2 years ago
7 0

Answer:

1. He or she can pedal harder to increase the speed of the bicycle.

2. He or she can apply the breaks to decrease the speed of the bicycle

3. He or she can turn the handle bars to change the bicycle's direction.

explanation:

:)

netineya [11]2 years ago
6 0

Answer:

describe three different ways to change your velocity when youre riding a bike?

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Please help and check all that apply and I will mark brainliest if it’s correct
Yuri [45]
A syncline is visable
3 0
3 years ago
Which statement best explains how isotopes can have different masses and still be the same element?
ss7ja [257]

Your list of choices is a very short list, and doesn't include any
correct explanation.

The mass of an atom is roughly the number of protons AND neutrons
in its nucleus, but the element only depends on how many protons are there. 

Different atoms of the same element may have different numbers of
neutrons, so their masses are different.  But they're the same element,
because they all have the same number of protons.

3 0
2 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

4 0
3 years ago
A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
2 years ago
You are traveling on an airplane. The velocity of the plane with respect to the air is 110.0 m/s due east. The velocity of the a
Mice21 [21]
1. Vpa = 180m/s. @ 0 deg.
  Vag = 40m/s @ 120 deg,CCW.


<span> Vpg = Vpa + Vag,
 Vpg = (180 + 40cos120) + i40sin120,
  Vpg = 160 + i34.64,
 Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
  A = 12.2 deg,CCW. = 12.2deg. North of East. </span>

3.  1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.

hope this helps</span>
8 0
2 years ago
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