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Dmitry [639]
3 years ago
12

When dropped from the same height, will a minivan or tricycle hit the ground first? Why?

Physics
1 answer:
Serga [27]3 years ago
5 0

Minivan hits the ground first than tricycle due to heavier mass and lesser air resistance.

Explanation:

According to Galileo's motion of objects, when two objects having different masses and areas dropped from same height does not hit the ground at the same time.

it's because the air offers much greater resistance to the falling motion of the lighter object than it does to the heavier one. The air is actually an upward force of friction, acting against gravity and slowing down the rate at which the tricycle falls.

Due to this when a minivan or tricycle when dropped from same height hits the ground at different times. As friction of air offers greater resistance to tricycle. Since minivan is heavier there will be lesser impact of air on it. So, minivan hits the ground first than tricycle.

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Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 16 m to the right to
Oliga [24]

Answer:

a

The total distance is d_t =19.1 m

b

The displacement is

  D_t = 12.9m

Explanation:

From the question we are told that

        Distance traveled by the ball for first player d_f = 16m to the right

        Distance  traveled by the ball for second player d_s = 3.1 m to the left

       

The total distance traveled by the ball is mathematically represented as

                   d_t = d_f + d_s

Substituting values

                  d_t = 16 + 3.1

                  d_t =19.1 m

The displacement is mathematically represented as

              D_t  = d_f -d_s

This is because displacement deal with direction and from the question we are told that right is positive and left is negative

          Substituting values  

                D_t = 16 -3.1

                 D_t = 12.9m

     

5 0
3 years ago
Describe the motion above and answer
amm1812

Answer:

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Explanation:

mmmmmmmmmmmmmmmmmmmm

7 0
3 years ago
A billiard ball traveling at 3.4 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at re
Tpy6a [65]

Answer:

Explanation:

Given

Initial velocity of first billiard ball u_1=3.4\ m/s

Initial velocity of second billiard ball u_2=0\ m/s

After collision first ball comes to rest

suppose m is the mass of both the balls

Conserving momentum to get the speed of second ball after collision

Initial momentum P_i=mu_1+mu_2=m\cdot 3.4+0

Final momentum P_f=mv_1+mv_2

where v_1 and v_2 are the speed of first and second ball respectively

P_i=P_f

m\cdot 3.4+0=0+m\cdot v_2

v_2=3.4\ m/s

thus speed of second ball after collision is equal to speed of first ball before collision

3 0
3 years ago
Read 2 more answers
Two point charges, A and B, are separated by a distance of 22.0 cm . The magnitude of the charge on A is twice that of the charg
REY [17]

Answer:

1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

Explanation:

From coulomb's law,

F = kAB/r².............................. Equation 1

Where F = Force exerted by each charge, A = charge at point A, B = charge at point B, r = distance of separation between the points, k = constant of proportionality.

Given: F = 47 N, r = 22 cm = 0.22 m.

Constant: k = 9.0×10⁹ Nm²/C²

Let: B = q, the A = 2q.

Substituting these values into equation 1,

47 = 9.0×10⁹(q×2q)/0.22²

47 = 18×10⁹(q²)/0.0484

q² = (47×0.0484)/(18×10⁹)

q² = 0.126×10⁻⁹

q² = 1.26×10⁻¹⁰

q = √( 1.26×10⁻¹⁰)

q = 1.12×10⁻⁵ C

The charge at point A = 2q = 2× 1.12×10⁻⁵  = 2.24×10⁻⁵ C.

Hence the charges are 1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

3 0
3 years ago
Please help! Will give brainly, 50 points!! I don't understand! I just need this to give me a boost on my other questions.
harina [27]

Answer:

The answer to your question is: c = 353 mi/h; Ф = 7.3°

Explanation:

You can see the picture below

To solve this problem, just imagine a right rectangle which the legs is 350 mi/h and 45 mi/h and we are going to find the hypotenuse.

                    c² = a² + b²

                    c² = 350² + 45²

                   c² = 122500 + 2025

                   c² = 124525

                   c = 352.9 ≈ 353 mi/h

tanФ = 45/ 350

tan Ф = 0.12

Ф = 7.3°

8 0
3 years ago
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