One disadvantage of cloning is

When an ice cube is placed in a glass of warm water, how are the signs and values q for the ice and the warm water related, assu

Valentin [98]

Answer:

q(ice) = -q(warm water)

Explanation:

Ice removes energy if ice is put in warm water and ice retains the very same amount of energy. Thus the water temperature heat sign is negative, and the ice heat sign is positive.

3 0

3 years ago

Rock salt contains sand and salt. A student mixed rock salt in warm water. Describe how the student would separate out the sand,

geniusboy [140]

Answer:

Explanation:

First, he can add the rock salt to the water and mix it well. Then he can remove the sand by decantation, after the sand has sedimented. He can seperate the water from the salt by evaporation. He should evaporate the water completely and capture the steam. the steam will cool into water. After evaporating completely, the salt will remain in the container

4 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago

To drive to the store from home, somebody first travels 5 km west, then they travel 5 km northwest. What is the straight line di

schepotkina [342]

Straight line distance between their home and the store can be solve using cosine law. first is solve the angle which is180 - 45 = 135 degree
C^2 = a^2 + b^2 - 2ab cos(135)c^2 = 5^2 + 5^2 - 2(5)(5) cos(135)c^2 = 85.35c = 9.24 km is the straight line distance between their home and the store

6 0

3 years ago

Based on the Law of Conservation of Matter, what number should be placed in the box to balance the chemical equation below?

Rashid [163]

Answer:

2

Explanation:

Step 1: RxN

4Al + 3O₂ → Al₂O₃

Step 2: Balance RxN

We have 4 Al's and 6 O₂'s on the reactant side.

We need the same number of Al's and O₂'s on the product side.

4Al + 3O₂ → 2Al₂O₃

7 0

3 years ago

Read 2 more answers