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Svetach [21]
2 years ago
14

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A littl

e later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:
a. it has the same position and the same velocity as A.b. it has the same position and the same acceleration as A.c. it has the same velocity and the same acceleration as A.d. it has the same displacement and the same velocity as A.e. it has the same position, displacement and velocity as A.
Physics
1 answer:
ICE Princess25 [194]2 years ago
3 0

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

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The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
2 years ago
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A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions
erastova [34]
•wind
•snow
•high tide/low tide
•thunder/lightning storms
5 0
2 years ago
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
1 year ago
An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
Aneli [31]

We are given: Final velocity (v_f)=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


5 0
2 years ago
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
Read 2 more answers
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