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Natali [406]
2 years ago
14

A block of mass M is attached to a spring of negligible mass and can slide on a horizontal surface along the x-direction, as sho

wn above. There is friction present between the block and the surface. The spring exerts no force on the block when the center of the block is located at x=0. At the instant shown above, the block is located at x=0 and moving toward the right with speed v=v0. (a) For the instant shown above, with the block located at x=0 and moving to the right, predict the direction of the net force on the block. If the net force is zero, select “The net force is zero.”
_ To the left _ The net force is zero _ To the right
Briefly justify your prediction.
Physics
1 answer:
Hoochie [10]2 years ago
6 0

Answer:

Because of the frictional force, the net force will oppose direction of the block and be directed towards the left even tho the spring exerts no force at this point

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Air pressure decreases with altitude

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Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts
Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

v=u+at

Substitute the values

10=0+162a=162a

a=\frac{10}{162}=0.0617m/s^2

s=ut+\frac{1}{2}at^2

Substitute the values

s=\frac{1}{2}(0.0617)(162)^2=809.6 m

s'=vt'=10\times 822=8220 m

a'=\frac{v}{t''}=\frac{10}{246}

s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0
3 years ago
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A taxi is travelling at 15m/s. Its driver accelerates with acceleration 3m/s^2 for 4 s. What would be the new velocity?..... pls
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Answer:

27 m/s

Explanation:

Given:

v₀ = 15 m/s

a = 3 m/s²

t = 4 s

Find: v

v = at + v₀

v = (3 m/s²) (4 s) + (15 m/s)

v = 27 m/s

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3 years ago
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Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

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K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
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