Every summer I drive to Pennsylvania. It is 895 km to get there. If I average 100 km/hr, how much time will I spend driving?

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260

ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

$\tau=F\times r$

$I\times\alpha=F\times r$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r$

$F=\dfrac{M}{2}a$ ...(I)

Here, F = tension

The force is

$F=m(g-a)$ ...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

$\dfrac{M}{2}a=m(g-a)$

$a=\dfrac{g}{1+\dfrac{M}{2m}}$

Put the value into the formula

$a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}$

$a=6.84\ m/s^2$

We need to calculate the tension in the rope

Using equation (I)

$F=\dfrac{M}{2}a$

Put the value into the formula

$F=\dfrac{12.1}{2}\times6.84$

$F=41.38\ N$

Hence, The tension in the rope is 41.38 N.

6 0

2 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

$V_f^2-V^2=-2gh$

$V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m$

$V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2$

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

$V_f=42.5m/s$

3 0

2 years ago

How to solve arctan[tan(7pi /4)] ...?

Svet_ta [14]

Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :

arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

2 years ago

Question 6 of 10

user100 [1]

Answer:

the answer is c

Explanation:

8 0

2 years ago

Read 2 more answers

A girl and a boy pull in opposite directions on strings attached to each end of a spring balance. Each child exerts a force of 2

BabaBlast [244]

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO generally: (left or down) would be considered negative in an equation. And the other person (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

0 is the number of equilibrium.

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

5 0

2 years ago