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skad [1K]
4 years ago
10

Where are chromosomes found in an animal cell?

Physics
2 answers:
timama [110]4 years ago
7 0
Chromosomes<span> are thread-like structures </span>located<span> inside the nucleus of </span>animal<span> and plant </span>cells<span>.</span>
anygoal [31]4 years ago
6 0
Inside the nucelus. it may b indifferent places fro other things but for animals it is in the nucleus 


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The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2
Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0
3 years ago
Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.
GalinKa [24]

Answer:

The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is equal to 2 : 1

Explanation:

Detailed explanation and calculation is shown in the image below

8 0
3 years ago
If the temperature of the metal oxide and water solution is increased,this would most likely..
kiruha [24]
Chemical Reaction between metal oxide and water solution
7 0
3 years ago
An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the
Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

x = v_0 t \frac{1}{2} at^2

Where,

x= Displacement

v_0 = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

L = \frac{1}{2} a t_1 ^2

For the second cart

2L \frac{1}{2} at_2^2

When the tenth car is aligned the length will be 9 times the initial therefore:

9L = \frac{1}{2} at_3^2

When the tenth car has passed the length will be 10 times the initial therefore:

10L = \frac{1}{2}at_4^2

The difference in time taken from the second car to pass it is 5 seconds, therefore:

t_2-t_1 = 5s

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

\frac{1}{2} = (\frac{t_1}{t_2})^2

t_1 = \frac{t_2}{\sqrt{2}}

From the relationship when the car has passed and the time difference we will have to:

(t_2-\frac{t_2}{\sqrt{2}}) = 5

t_2 (\sqrt{2}-1) = 3\sqrt{2}

t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Replacing the value found in the equation given for the second car equation we have to:

\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Finally we will have the time when the cars are aligned is

18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2

t_3 = 36.213s

The time when you have passed it would be:

20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2

t_4 = 38.172

The difference between the two times would be:

t_4-t_3 = 38.172-36.213 \approx 2s

Therefore the correct answer is C.

4 0
3 years ago
Compare skater's total energy at point A and at point E?(disregard the friction)
White raven [17]

Answer:

c

Explanation:

because I've had this question before and got it right

7 0
3 years ago
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