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soldi70 [24.7K]
3 years ago
7

a 300-gram toy car and a 500-gram toy car are sitting on a shelf that is 2 meters highter than the floor. by what percent is the

PE of the 500-gram car grater than the PE of the 300-gram car?
Physics
2 answers:
andre [41]3 years ago
7 0
We are given:

m1 = 300 grams
m2 = 500 grams
h1 = h2 = 2 meters

We have to determine the potential energy of both toy cars:

PE = mgh
PE1 = m1gh1
PE1 = 300 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE1 = 5.886 Joules

PE2 = m2gh2
PE2 = 500 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE2 = 9.81 Joules

PE2 - PE1 / PE2

9.81 - 5.886 Joules / 9.81 joules * 100 = 40.0%

Therefore, the 500-gram car is greater by 40% in terms of potential energy than the 300-gram car. This is because mass is directly proportional to the potential energy of an object. 

White raven [17]3 years ago
4 0

Answer:

66.67% of 500 g Potential energy is greater than that of 300 g of Potential energy.

Explanation:

Height of 300 g mass=h_{1}= 2 m

Height of 500 g mass=h_{2}=2 m

h_{1}=h_{2}

and also gravity remains constant

hence

% of potential energy of 500 g greater than of 300 g=((500-300)/300)*100

% of potential energy greater=(2/3)*100

% of potential energy greater=66.67%  

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vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

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The resulting nucleus is Tc.

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4 0
3 years ago
An airplane during departure has a constant acceleration of 3 m / s².
Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>

t

3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>

8

24 = v - 20

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6 0
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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
In-s [12.5K]

Answer:

 v = -v₀ / 2

Explanation:

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            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

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          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

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