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Troyanec [42]
3 years ago
9

A horse draws a sled horizontally across a snow-covered field. The coefficient of friction between the sled and the snow is 0.13

5, and the mass of the sled, including the load, is 195.9kg. If the horse moves the sled at a constant speed of 1.785m/s, what is the power needed to accomplish this?
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

P = 462.62 watts

Explanation:

The power needed to accomplish this can be calculated how

P = Fv

    Where F:  The force exerted by the horse

                v:  velocity

The force exerted by the horse is against friction force; how the movement is with constant velocity these forces must be equals, then

Fr = μN

    =μmg

    = (0.135)(195.9)(9.8)

    = 259.17 N

And the power is

P = (259.17)(1.785)

P = 462.62 watts

You might be interested in
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed
Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0
2 years ago
The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli
Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

4 0
2 years ago
wo plates with area 6.00×10−3 m2 are separated by a distance of 1.50×10−4 m . If a charge of 5.40×10−8 C is moved from one plate
liq [111]

Answer:

V = 152.542 volts

Explanation:

Given data:

area of plates6.00\times 10^{-3} m^2

distance between the plates is 1.50\times 10^{-4} m

charge = 5.40\times 10^{-8} c

we know that capacitance is given as

C = \frac{\epsilon A}{d}

C = \frac{8.85\times 10^{-12} 6\times 10^{-3}}{1.50\times 10^{-4}}

C = 3.54\times 10^{-10} F

potential difference is given as

V =\frac{Q}{C} = \frac{5.40\times 10^{-8}}{3.54\times 10^{-10}}

V = 152.542 volts

3 0
3 years ago
A 3.04 kg particle is located on the x-axis at xm = −8 m, and a 5.61 kg particle is on the x-axis at xM = 3.56 m. Find the coord
Murljashka [212]

Answer:

center of mass = −0.50 m

Explanation:

given data

mass m1 = 3.04 kg

distance xm = -8 m

mass m2 = 5.61 kg

distance xM = 3.56 m

solution

we get here center of mass for n mass of system that is express as

center of mass = \frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}     ......................1

but we have only 2 particle system so we will get

center of mass = \frac{m1 \times xm+m2 \times xM}{m1+m2}      .................2

put here value and we will get

center of mass = \frac{3.04 \times (-8 )+5.61 \times 3.56}{3.04 + 5.61}

solve it we will get

center of mass = −0.50 m

8 0
3 years ago
The diagram below represents the rock cycle.
lara31 [8.8K]

Answer:

C. Heat and Pressure

Explanation:

The arrow which is labeled A points from igneous rock to metamorphic rock.

There are three types of Rocks:

1. Igneous Rock

2. Metamorphic Rock

3. Sedimentary Rock

Rock cycle:

Rock cycle is the process that describes the transition between these three types of rocks. Each type has its own form and its own equilibrium condition. The rock type alters when it is pushed out of its equilibrium conditions.

Transition of Igneous rock to Metamorphic rock:

Igneous rock forms when magma cools down. The transition of Igneous Rock to Metamorphic Rock is a result of a process called Metamorphism. Metamorphism is the alteration in the structure of rock as a result of certain heat and pressure conditions. Inside Earth heat comes from pressure. Heat with pressure does not melt the rock but it bakes the rock. Baking is not melting but it changes the shape of the rock while it is still solid. It actually forms crystals. Because the rock changes its structure, it is called Metamorphic Rock.

5 0
3 years ago
Read 2 more answers
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