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Alchen [17]
3 years ago
8

Which of the following is a mixture: a. air c. water b. salt d. sulfur

Physics
1 answer:
murzikaleks [220]3 years ago
3 0
A. Air because is composed of different elements such as nitrogen, oxygen, water vapor, carbon dioxide, sulfur, argon and many others.
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List three devices that use electric current from batteries and three that use regular house current.
Bond [772]
Charger
counsels
TV

Fan
light
car
4 0
4 years ago
Consider the following:
mihalych1998 [28]

Answer:

I think it's 3 c only the loudness of the sound

7 0
3 years ago
What is the atomic number of an atom?
Sever21 [200]

Answer:

B, the number of protons.

Explanation:

It is the number of protons found in the nucleus of the atom.

4 0
3 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
When is a zero not significant?
maria [59]

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0
4 years ago
Read 2 more answers
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