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sdas [7]
3 years ago
8

Peter throws a snowball at his car parked in the driveway. The snowball disintegrates as it hits the car. By Newton’s third law,

you know that both the car and the snowball exert a force on each other. What can you say about the magnitude of the forces exerted by the snowball and the car?
Physics
1 answer:
elena55 [62]3 years ago
3 0
Well, they are equal and opposite, but the force is high enough to break up the snowball.  That is where the energy is dissipated.
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. A toy rocket has a mass of 350 g at launch. The force it produces
nydimaria [60]

Answer:

the acceleration is reduced by gravity

a = (15 / .35) - [9.8 * sin(65º)]

Explanation:

break the launch vector into two components, vertical and horizontal

Force Net Vertical=-9.8*.350+15cos65 N

force net horizonal=15sin65

initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i

using i,j vectors..

5 0
3 years ago
How does the speed of the different molecules that make up the air depend on their masses?
koban [17]
Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?

8 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F  = (1200)(0) = 0

4 0
3 years ago
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