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2 years ago

A 75.0 kg sailor climbs a 28.3 m rope ladder at and angle of 45.0degrees with the mast. how much work did he do?

2 answers:

8 0

W = m · g · h
h = 28.3 m · sin 45° = 28.3 m · 0.707 = 20 m
g = 9.8 m/s²
W = 75 kg · 9.8 m/s² · 20 m
Answer:
W = 14,700 J = 14.7 kJ

crimeas [40]2 years ago

7 0

Explanation :

It is given that,

mass of sailor, m = 75 Kg

length of rope, h = 28.3 m

work done, $w=Fdcos\theta$

$w=mghcos\theta$

$w=75\ Kg\times 9.8\ m/s^2\times 28.3\ m\ cos(45)$

$w=10,816\ J$

$w=10.816\ kJ$

Hence, it is the required solution.

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-A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back

Llana [10]

Explanation:

Eg = mgh

a. Eg = (2.00 kg) (9.8 m/s²) (12.0 m)

Eg = 235 J

b. Eg = (2.00 kg) (9.8 m/s²) (3.00 m)

Eg = 58.8 J

c. Eg = (2.00 kg) (9.8 m/s²) (4.50 m)

Eg = 88.2 J

d. When the coconut hits the bystander:

Ek = 235 J − 58.8 J = 176 J

When the coconut hits the ground:

Ek = 88.2 J − 0 J = 88.2 J

Ek is the greatest when the coconut hits the bystander.

3 0

2 years ago

A superman cyclist rode a bike uphill at 20 miles/hour for two hours. To sustain this constant speed the cyclist was exerting 50

NeX [460]

Answer:

1.056 x 10⁷ lb-ft

Explanation:

v = Speed of the bike = 20 mph

t = time of travel = 2 h

d = distance traveled by cyclist

Distance traveled by cyclist is given as

d = v t

d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

F = force applied by cyclist = 50 lb

W = work done by cyclist

Work done by cyclist is given as

W = F d

W = (50) (211200)

W = 1.056 x 10⁷ lb-ft

5 0

2 years ago

Charlie is playing with his daughter Torrey in the snow. She sits on a sled and asks him to slide her across a flat, horizontal

professor190 [17]

Answer:

Explanation:

Pulling up with the rope would decrease the frictional force, pushing down would increase the frictional force.

3 0

2 years ago

Read 2 more answers

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

2 years ago

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

-Dominant- [34]

The car would go from zero to 58.0 mph in 2.6 sec.

Since the force on the car is constant, therefore the acceleration of the car would also be constant.

Now for constant acceleration we can use the equation of motion

Using first equation of motion to calculate the acceleration of the car

v=u+at

29=0+a×1.30 ...... Eq. (1)

Again using the first equation of motion

58=0+a*t ....... Eq. (2)

Dividing eq. (2) with equation 1

t=2×1.3

t=2.6 sec

7 0

2 years ago