What is an error in a lab mean?

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0

3 years ago

What are some thermal insulators you can find in a normal house

adoni [48]

Body heat :)

i'm going du.mb here

6 0

3 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

$p_f=p_i$

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

$p_{of}+p_{wf}=p_{oi}+p_{wi}$

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

$m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}$

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so $v_{oi}=v_{wi} = 0\frac{m}{s}$ :

$m_o*v_{of}+m_w*v_{wf}=0$

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for $v_{wf}$ :

$v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}$

$v_{wf}=-6.06\frac{m}{s}$

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0

3 years ago

An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y

Xelga [282]

Answer:

The fraction of the protons would have no electrons $=1.88\times 10^{-10}$

Explanation:

We are given that

Amoeba has total number of protons= $1.00\times 10^{16}$

Net charge, Q=0.300pC

Electrons are fewer than protons= $1.88\times 10^6$

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

= $\frac{Fewer\;electrons}{Total\;protons}$

The fraction of the protons would have no electrons

= $\frac{1.88\times 10^{6}}{1.00\times 10^{16}}$

$=1.88\times 10^{-10}$

Hence, the fraction of the protons would have no electrons $=1.88\times 10^{-10}$

6 0

3 years ago

Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv

deff fn [24]

Answer:

electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

λ = q / x

As it is constant, we can write it based on differentials

λ = dq / dx

dq = λ dx

We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

E = k ∫ λ dx / x²

E = k la (- 1 / x)

Let's get the negative sign from the parentheses

E = - k λ (1 / x)

E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

E = - k Q/a [a / r (r-a)]

E = - k Q (1 /r(r-a))

b) We calculate the force.

F = E qo

F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

F = - k Qqo / r² (1 - a/r)

F ≈ - k Q qo / r²

6 0

3 years ago