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3 years ago

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

1 answer:

6 0

How much work in J does the string do on the boy if the boy stands still?

answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero 
--------------------------------------... 
How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? 

answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. 

If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: 
4.5N + (boys negative acceleration * mass) = total force1 
work = total force1 x 11 meters 
--------------------------------------... 

How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? 

answer: same as above only reversed: 
4.5N - (boys negative acceleration * mass) = total force2 
work = total force2 x 11 meters

You might be interested in

sketch a graph of position vs. time for a person starting 0.4 m away and standing still. Include an explanation of your predicti

Korvikt [17]

Answer:

Straight line parallel to time axis.

Explanation:

The slope of the position time graph gives the velocity.

As the man is still, that means the velocity is zero. So, the slope of the graph is zero. It is a straight line parallel to time axis.

5 0

2 years ago

A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and

N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t where t is time

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0

3 years ago

Read 2 more answers

3. An astronaut on a space-walk swings a 0.5 kg metal tool in a circular path at the end of a cable 3.6 m long. The astronaut th

Semenov [28]

Answer:

Explanation:

2 km a hour

6 0

2 years ago

What the kinetic energy quantities in calculation pls help me

Rashid [163]

Answer:

KE = 0.5 * m * v², where: m - mass, v - velocity.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2.

3 0

2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago