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Readme [11.4K]
3 years ago
15

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

isplacement vector of the tip are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the (c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle for the hour after that? Give all angles as positive values measured counterclockwise from the +x direction (to the right, or 3 o'clock).
Physics
1 answer:
olya-2409 [2.1K]3 years ago
7 0

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

|r|=\frac{3\times 2\pi r}{4}=75.408 cm

Angle =270^{\circ}

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

|r|=\frac{2\pi r}{4}=25.136 cm

angle turned =90^{\circ}

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. 90 ^{\circ}

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a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin
trapecia [35]

Answer:

a) E_{p} = 0

E_{k} = 168.7 J

E_{m} = 168.7 J

b) E_{p} = 73.6 J

E_{k} = 95.8 J

E_{m} = 169.4 J

c) E_{p} = 169.2 J

E_{k} = 0

E_{m} = 169.2 J

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

E_{p} = mgh = 0

The kinetic energy is:

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

E_{p} = mgh = 1.5*9.81*5 = 73.6 J

Now, to find the kinetic energy we need to calculate the speed at 5 m:

v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9

v_{f} = \sqrt{126.9} = 11.3 m/s

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J

c) At its maximum height:

v_{f}: is the final speed = 0

h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m

Now, the potential, kinetic and mechanical energies are:

E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J

E_{k} = \frac{1}{2}mv^{2} = 0

E_{m} = 169.2 J + 0 = 169.2 J

I hope it helps you!    

7 0
2 years ago
During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce
Mrrafil [7]

Answer:

When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).

In order to climb, you need to work against work done by gravity on the pack.

Hence work done by you = work done by gravity on pack  

                                       = Force x displacement = 70 x 30 = 2100 J.

 

So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.

Therefore power = work done by you/time(in seconds)

                           =     2100/600 =3.5 watts

Explanation:

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2 years ago
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A raised plaster cover (often called a plaster ring or mud ring) is permitted to increase the maximum number of conductors permi
andreev551 [17]

Answer:

When it is marked with its cubic-inch volume

Explanation:

Because this allows for best and efficient identification

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3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or
Vesnalui [34]

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

4 0
2 years ago
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