Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by
![F_s = \mu N](https://tex.z-dn.net/?f=F_s%20%3D%20%5Cmu%20N)
where
is the coefficient of friction and
is the normal force.
When the student pulls on the block with force
at an angle
, the normal force on the block becomes
![N = Mg- F_psin(\theta)](https://tex.z-dn.net/?f=N%20%20%3D%20Mg-%20F_psin%28%5Ctheta%29)
and hence the frictional force becomes
.
Now, as we increase
,
increases which as a result decreases the normal force
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
Answer
Around 400 B.C.E, the Greek philosopher Democritus introduced the idea of the atom as the basic building block matter. Democritus though that atoms are tiny, uncuttable, solid particles that are surrounded by empty space and constantly moving at random.
Pls give me BRAINLIEST
Answer:
a.![2.86 m/s^2](https://tex.z-dn.net/?f=2.86%20m%2Fs%5E2)
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N
![\mu_s=0.14](https://tex.z-dn.net/?f=%5Cmu_s%3D0.14)
![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
By Newton's second law
![8F-f=(8M+m)a](https://tex.z-dn.net/?f=8F-f%3D%288M%2Bm%29a)
![a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B8F-f%7D%7B8M%2Bm%7D%3D%5Cfrac%7B8%28185%29-%280.14%29%289.8%29%28250%29%7D%7B8%2818.5%29%2B250%7D)
![a=2.86 m/s^2](https://tex.z-dn.net/?f=a%3D2.86%20m%2Fs%5E2)
b.By Newton's second law
![T=ma+\mu_s mg](https://tex.z-dn.net/?f=T%3Dma%2B%5Cmu_s%20mg)
Substitute the values
![T=250\times 2.86+0.14(250)(9.8)=1058 N](https://tex.z-dn.net/?f=T%3D250%5Ctimes%202.86%2B0.14%28250%29%289.8%29%3D1058%20N)
Hence, the force in the coupling between the dogs and the sled=1058 N
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
![V(t) = \epsilon. e^{-t.\frac{L}{R} }](https://tex.z-dn.net/?f=V%28t%29%20%3D%20%5Cepsilon.%20e%5E%7B-t.%5Cfrac%7BL%7D%7BR%7D%20%7D)
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
![0.8*19 = 19. e^{-4.\frac{22}{R} }](https://tex.z-dn.net/?f=0.8%2A19%20%3D%2019.%20e%5E%7B-4.%5Cfrac%7B22%7D%7BR%7D%20%7D)
![0.8 = e^{-\frac{88}{R} }](https://tex.z-dn.net/?f=0.8%20%3D%20e%5E%7B-%5Cfrac%7B88%7D%7BR%7D%20%7D)
![ln(0.8) = ln(e^{-\frac{88}{R} })](https://tex.z-dn.net/?f=ln%280.8%29%20%3D%20ln%28e%5E%7B-%5Cfrac%7B88%7D%7BR%7D%20%7D%29)
![ln(0.8) = -\frac{88}{R}](https://tex.z-dn.net/?f=ln%280.8%29%20%3D%20-%5Cfrac%7B88%7D%7BR%7D)
![R = -\frac{88}{ln(0.8)}](https://tex.z-dn.net/?f=R%20%3D%20-%5Cfrac%7B88%7D%7Bln%280.8%29%7D)
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.