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Rudik [331]
2 years ago
10

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°

to the ground. How much force did the bulldozer apply to the mound of soil? Round your answer to two significant figures.
Physics
2 answers:
qaws [65]2 years ago
7 0

Answer:

520

Explanation: answer on edge

spin [16.1K]2 years ago
3 0

<em>Answer</em>


Force = 170 N



<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.


Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m


Now;

work done = force × distance.


4500 J = force × 26.15


dividing both sides by 26.15,


Force = 4500/26.15

= 172.07 N


Answer to two significant figures = 170 N

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The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who
densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

L = \frac{y \times d}{1.22 \times  \lambda} = \frac{1.2 \times 0.004}{1.22 \times  400 \times 10^{-9}} = 9836.066 \ km

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)  

= 1.22 \times \frac{400 \times 10^{-9}}{0.004}

sinθ = 1.22×10⁻⁴ rad

6 0
2 years ago
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For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat
andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

B(t)=5t\ T, t is time in seconds

\dfrac{dB}{dt}=5\ T

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

E(2\pi x)=\dfrac{d\phi}{dt}

x is the distance from the axis of the solenoid

E(2\pi x)=\pi r^2\dfrac{dB}{dt}, r is the radius of the solenoid

r^2=\dfrac{2xE}{(dE/dt)}

r^2=\dfrac{2\times 2\times 1.1}{(5)}

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0
2 years ago
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A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe
natta225 [31]
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
4 0
2 years ago
When boiling water, a hot plate takes an average of 8 minutes and 55 seconds to boil 100 milliliters of water. Assume the temper
alexandr1967 [171]

Answer:

90.9 seconds

Explanation:

m = Mass of liquid = Volume×Density

c = Specific heat

\Delta T = Change in temperature

t = Time taken

Room temperature = 75 °F

Converting to Celsius

(75-32)\times \frac{5}{9}=23.889\ ^{\circ}C

Heat required to raise the temperature of water

Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 1000\times 4186\times (100-23.889)\\\Rightarrow Q=31860.0646\ J

Power

P=\frac{Q}{t}\\\Rightarrow P=\frac{31860.0646}{8\times 60+55}\\\Rightarrow P=59.55152\ W

Efficiency of the plate

\frac{59.5512}{283}\times 100=21.04282\%

Heat required to raise the temperature of water

Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 784\times 2150\times (56-23.889)\\\Rightarrow Q=5412.63016\ J

P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{5412.63016}{0.2104282\times 283}\\\Rightarrow t=90.9\ s

Time taken to heat the aceton is 90.9 seconds

4 0
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a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being
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From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

  35 ft + 27 ft + 27 ft + 18 ft = 107 ft

3 0
2 years ago
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