How can a cyclist's acceleration change even if its velocity remains constant?

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express

Tatiana [17]

Answer:

at x=0

$T_{l}$ =W*L/L

and

$T_{l}$ =W- $T_{r}$

Explanation:

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

moments is the product of force and the perpendicular distance in line of the action of the given force

from the principle of moments which states that the sum of clockwise moments ,must be equal to the sum of anticlockwise moments.

also, sum of upward forces must be equal to sum of downward forces

Going by the aforementioned,

taking moments about $T_{r}$

W*(l-x)= $T_{l}$ *(L)

$T_{l}$ =W*(L-x)/(L)..............1

at x=0

$T_{l}$ =W*L/L

also

$T_{l}$ + $T_{r}$ =W

$T_{l}$ =W- $T_{r}$

3 0

3 years ago

What is the velocity of a 12kg object that has the momentum of 60kgm/s?

Bad White [126]

Answer: 5 m/s

Explanation:

Momentum = mv (mass*velocity)

60kg*m/s = (12kg)*(v)

v = (60kg*m/s)/(12kg)

v = 5 m/s

4 0

3 years ago

Jason drives uphill from home to school, averaging 20 miles per hour. On the way back, he drives back downhill from school to ho

r-ruslan [8.4K]

Answer:

Average velocity = 30 miles per hour

Explanation:

Given that

Velocity for uphill = 20 miles per hour

Velocity for downhill = 60 miles per hour

We know that

Distance = Velocity x time

Total distance = Average velocity x total time

Lets take distance from home to school is s

Time taken from home to school= s/20 hr

Time taken from school to home= s/60 hr

Total time =s/20+s/60

Total distance = Average velocity x total time

s + s = Average velocity x (s/20+s/60)

So average velocity = 30 miles per hour

6 0

3 years ago

Please urgent science question explaining needed

sweet-ann [11.9K]

Answer:

what is the question? there isn't one

3 0

3 years ago

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