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3 years ago

Two skaters begin at rest. Skater 1 has a smaller mass than Skater 2. After they push off each other,...

Physics

2 answers:

Vladimir [108]3 years ago

6 0

Answer:

C. Sakter 1 has Greater velocity than skater 2.

Explanation:

This concept can be explained using the concept of collision between two particles with different masses. When both the masses pull each other, smaller skater will acquire the greater velocity due to its less mass. While the greater skater will have lower velocity due to its greater mass. Hence, C is the right answer.

As the momentum is the product of mass and velocity, therefore change in momentum in both the skaters will be equal. It is obvious that both skaters have unequal masses and velocity but the product of mass and velocity for them is equal. Hence, option A and B are not right answers.

When both skaters pull each other, they exert force on each other. According to newton’s third law of motion action and reaction forces are equal and are in opposite direction. So, the forces which both skaters will experieces are equal. Hence, D is also not a right answer.

klasskru [66]3 years ago

3 0

As both the skater pushed each other with a force, skater 1 will acquire greater velocity than skater 2 with heavier mass.

Answer: Option C

Explanation:

Here we can understand the scenario with the example of a particle collision. When two particles of different masses, one being lighter and the other being heavier, there will be a movement and change in velocities.

Now, when these two particle collide with each other, there will be an equal and opposite thrust or reaction on both the particles. The particle with heavier mass will not be able to acquire more velocity because of its weight whereas the lighter particle will be pushed far away, being less in mass.

We can also say that the heavier particle will exert more force as compared to the lighter one and hence, can easily push it far away. this is all because the force is directly proportional to the mass of the object.

Hence, just like the particle, when two skater try to push each other with a significant amount of force, skater 2 will win to exert more force and therefore, skater 1 will acquire greater velocity as compared to skater 2.

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Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i

OlgaM077 [116]

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

3 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid

ioda

Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116

4 0

2 years ago

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago

The dimension line is connected to the part being measured by _______________. A. Hidden lines B. Extension lines C. Visible lin

kondaur [170]

B: Extension Lines! You could have just searched this up on google

3 0

3 years ago

Read 2 more answers

Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0

2 years ago