All categories

3 years ago

HELPP!! what is not true about mitosis??

Physics

2 answers:

Andreyy893 years ago

7 0

Answer:

I think its A

Explanation:

Marina86 [1]3 years ago

5 0

I also think it’s A I just learned this

You might be interested in

______ is the total distance traveled divided by the total time of travel.

sweet-ann [11.9K]

Average speed is the answer

4 0

2 years ago

A 70-kg astronaut is space walking outside the space capsuleand is stationary when the tether line breaks. As a means of returni

Bogdan [553]

Answer:

0.4 m/s

Explanation:

Law of conservation of momentum tell us that the change in momentum of the hammer will be equal to the change in momentum of the astronaut

change in momentum of hammer = change in momentum of astronaut

2 kg (14 m/s - 0 m/s) = 70 kg * (v-0)

v = 0.4 m/s

7 0

2 years ago

Read 2 more answers

al aplicar una fuerza de 2 N sobre un muelle este se alarga 4cm.¿cuanto se alargara si la fuerza es el triple?¿que fuerza tendri

3241004551 [841]

1) 12 cm

2) 3 N

Explanation:

The relationship between force and elongation in a spring is given by Hooke's law:

$F=kx$

where

F is the force applied

k is the spring constant

x is the elongation

For the spring in this problem, at the beginning we have:

$F=2 N$

$x=4 cm$

So the spring constant is

$k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm$

Later, the force is tripled, so the new force is

$F'=3F=3(2)=6 N$

Therefore, the new elongation is

$x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm$

In this second problem, we know that the elongation of the spring now is

$x=6 cm$

From part a), we know that the spring constant is

$k=0.5 N/cm$

Therefore, we can use the following equation to find the force:

$F=kx$

And substituting k and x, we find:

$F=(0.5)(6)=3 N$

So, the force to produce an elongation of 6 cm must be 3 N.

6 0

3 years ago

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

4 0

2 years ago

A spring balance used to weigh candy is built with a spring the spring stretches 3.00 cm when a 15 n weight is placed in the pan

ira [324]

Answer: 5.9cm

Explanation:

15N/3cm=5

Convert 3kg to N (29.42N)

Use the 5 from earlier to divide 29.42/5=5.884

Round to 5.9

5.9cm is the answer

(I don’t know if I solved this the correct way but I got it right. 5.9 is the correct answer)

6 0

3 years ago