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dmitriy555 [2]
3 years ago
11

Mateo and Jackie were on a playground rolling some different-sized balls to get them to crash. They used a tennis ball (the same

mass) in both crashes. They tried two soccer balls—a pink one (more mass) for Crash 1 and a blue one (less mass) for Crash 2.
Mateo and Jackie want to know what happened to the tennis ball. Use information from the diagram to answer.


In which crash did the tennis ball experience a stronger force? How do you know?



a

Crash 1; the force on the soccer ball was stronger in this crash, so the force on the tennis ball was also stronger.


b

There was no force on the tennis ball. In each crash, only the soccer ball experienced a force.



c

It was the same in both crashes. The soccer ball changed speed by the same amount in each crash, so the force on the tennis ball was the same both times.



d

The diagram doesn’t tell you anything about the force on the tennis ball. It only gives information about the force on the soccer balls.

Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

In crash one, because the pink soccer ball had more mass, so it would have a stronger force.

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On which planet can you jump about the same height as Earth? Explain your answer.
Debora [2.8K]

Answer:

Venus and Earth

Explanation:

The force of gravity depends on your mass AND the mass of the planet you stand on. Earth and Venus are about the same size, so they have about the same mass. If you go to Venus, your mass hasn't changed, and the planet mass is almost the same as earth- so the force of gravity on you (AKA your weight) will be the same.

8 0
3 years ago
Read 2 more answers
2. Below are four velocity vs. time graphs. Which graph represents the motion of an object that
Airida [17]

Answer:

Answer (b)

Explanation:

Acceleration is the rate of change of velocity in time, therefore constant acceleration will be changing velocity in direct proportion to time and will have a linear plot on a velocity/time chart. This eliminates answer (d) which has an increasing acceleration in time

answer (a) has a negative slope so the acceleration would be considered negative

answer (c) has zero slope so acceleration is zero and velocity is constant.

answer (b) has the required positive slope and acceleration

6 0
3 years ago
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
When you look at a transverse wave, the distance from the baseline to the crest of the wave is called the wave's ___________.
uranmaximum [27]
Amplitude


Explanation have a great that
6 0
3 years ago
If a rock weighing 2,200 N is dropped from a height of 15 m, what is its kenetic energy just before it hits the ground
ycow [4]

Kinetic energy of the rock just before it hits the ground=KE=33000 J

Explanation:

Weight= 2200N

mg=2200

m(9.8)=2200

m=224.5 kg

initial velocity=0

final velocity =V

using kinematic equation V²=Vi²+2gh

V²=0+2 (9.8)(15)

V=17.1 m/s

now kinetic energy= 1/2 mV²

KE= 1/2 (224.5)(17.1)²

KE=33000 J

Thus the kinetic energy of the rock just before it hits the ground=33000 J

7 0
3 years ago
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