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Pachacha [2.7K]
3 years ago
6

The cold virus causes disease when it enters the _____ of the human body.

Physics
1 answer:
belka [17]3 years ago
7 0
Cells of the human body
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What is meant by the statement,the linear expansivity of copper is 0.000017k
Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

4 0
3 years ago
Jason wanted to find the Volume of two rocks How could you use the tools below that is shown to find the volume of these irregul
Marta_Voda [28]
I think the answer is  to measure the tubes
5 0
3 years ago
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In February 1955, a paratrooper fell 365 m from an airplane without being able to open his chute but happened to land in snow, s
DaniilM [7]

Answer:

a. i=4760 kg*m/s

b. D_U= 1.11 m

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

K_E= (1/2)*85kg*(56m/s)^2

K_E=133280 J

D_U= \frac{133280J}{120000N} = 1.11 m

b)

The momentum is equal to the impulse on him from the snow so:

p=m*v

p=85kg*56m/s

i=4760 kg*m/s

7 0
3 years ago
The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i
shtirl [24]

Answer: The coefficient of kinetic friction is μ = 0.6

Explanation:

For an object of mass M, the weight is:

W = M*g

where g is the gravitational acceleration: g = 9.8m/s^2

And the friction force between this object and the surface can be written as:

F = W*μ

where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)

In this case, the weight is:

W = 20N

And the friction force is:

F = 12N

Replacing these values in the equation for the friction force we get:

12N = 20N*μ

(12N/20N) = μ = 0.6

The coefficient of kinetic friction is μ = 0.6

7 0
3 years ago
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A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
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