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artcher [175]
3 years ago
5

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

Physics
1 answer:
Elenna [48]3 years ago
8 0
Let the ball is initially at point P1. Finally it reaches at point P2.( bottom most). Since only conservative force is acting on the ball throughout the motion, mechanical energy will be conserved. Therefore conserving energy from P1 to P2, Initially ball is at rest(P1). So initial kinetic energy is 0. Let ball is at height h1 initially, so initial potential energy = mgh ( m= mass of the ball). At bottom most point (P2), potential energy of ball changes into kinetic energy. So final kinetic energy =1/2 mv^2. Final potential energy = 0. Equating energies at P1 and P2, Mgh= 1/2 mv^2. Therefore, v= (2gh)^1/2. Since all the initial and final points are at same height, velocity will be same for all the paths(irrespective of the path taken by ball). So velocity will be same.
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What unit is used to count atoms and molecules?
murzikaleks [220]

the answer in my opinion would be A

7 0
3 years ago
A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

3 0
3 years ago
Acceleration increases over time once a force is applied to the object. Given a force of 10.0 Newtons, which mass will have the
Zolol [24]

Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...

A) 0.10 kg is lightest among them, so it's your answer

6 0
3 years ago
Read 2 more answers
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0
3 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
Drupady [299]

Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

Current in wire=40 A

Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

Substitute the values in the given formula

B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

B=4.5\times 10^{-5} T

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

7 0
3 years ago
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