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amid [387]
3 years ago
12

2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere

Physics
1 answer:
Roman55 [17]3 years ago
5 0

Answer:

yes

Explanation:

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a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c
ser-zykov [4K]
Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = \frac{7.4 - 7.6}{7.6} × 100

= 2.631 % = 2.7 % (approximately.)

hope it helps!

5 0
3 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
3 years ago
Crystal kick the 3.0 kg soccer ball horizontally off of the top of a building the building is 6.0 m tall and the ball lands 10.0
sattari [20]

Answer:

gametes

nucleic acids

Explanation:

amino acids

proteins

3 0
3 years ago
In a recent experiment, anti-bacterial hand sanitizer was compared to hand soap. Students washed their hands using the sanitizer
Elden [556K]
<span>B) Both soap brands out perform hand sanitizer in killing bacteria.</span>

Looking at the data, the conclusion is pretty obvious, the two different types of soap resulted in about half the bacteria cultures as either the hand sanitizer, or just using warm water. In fact, the hand sanitizer looks like it's even worse than simply using warm water. So looking at the four available options, the best choice is "B) Both soap brands out perform hand sanitizer in killing bacteria."
7 0
3 years ago
Read 2 more answers
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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