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nikdorinn [45]
2 years ago
10

a force of 17 acts on a block of mass 4kg the forces of friction opposing the motion is 5n . the acceleration of the block is​

Physics
1 answer:
Nady [450]2 years ago
8 0
Fk = μK N

N = m a

N = 4 × a

N = 4a

Fk = μK N

17 = 5 × 4a

17 = 20 a

a = ¹⁷/₂₀ = 0.85 m/s²
You might be interested in
If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no
Yakvenalex [24]

<u>Answer</u>

3.7 Km south


<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

                    = - 3.7 Km      But negative was towards south.

∴ Answer = 3.7 Km south.


6 0
3 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
Explain why, in terms of forces, there is a risk of head injury when diving from this height. Suggest why the high divers would
podryga [215]

Answer:

Due to lower risk of injury or damage.

Explanation:

The high divers would choose to enter the water from the feet first because there is low risk of injury. The brain is the most important part of the body which very sensitive to any small injury. Small injury to brain leads to big problems in life. High divers can reach speeds of nearly 60 mph and enters about 28m into the water in about three seconds which can damage the head region if comes in contact with the ground so this is the reason the high divers avoid of entering in the water through their heads and choose entering through their feet.

4 0
3 years ago
A race car has a centriple acceleration of 15.625 m/s as it goes around a curve if the curve is a circle with radius 40 m what’s
Elanso [62]

Answer:

25 m/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius.

a = v² / r

15.625 m/s² = v² / (40 m)

v² = 625 m²/s²

v = 25 m/s

The speed of the car is 25 m/s.

7 0
3 years ago
A baseball player throws a ball horizontally at the same moment another baseball player jobs a second ball straight down from th
attashe74 [19]

I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.

I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball <u><em>drop</em></u>, at the same time and from the same height as the first ball.

Now I'll go ahead and answer the question that I've just invented:

Strange as it may seem, <em>both</em> balls hit the ground at the <em>same time</em> ... the one that's thrown AND the one that's dropped.  The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.

And here's another example of the exact same thing:

Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else <em>drops</em> another bullet at exactly the same time, from a point right next to the end of the rifle barrel.  I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.

7 0
3 years ago
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