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valina [46]
3 years ago
11

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

the sled and surface is 0.229. What is the speed of the sled after it has traveled a distance of 3.89 m?
Physics
1 answer:
Darya [45]3 years ago
8 0

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

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Answer:

b. 600,000 J

Explanation:

Applying the law of conservation of energy,

The thermal energy created = Kinetic energy of the suv.

Q' = 1/2(mv²)............... Equation 1

Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.

From the question,

Given: m = 3000 kg, v = 20 m/s

Substitute these values into equation 1

Q' = 1/2(3000×20²)

Q' = 600000 J

Hence the right option is b. 600,000 J

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3 years ago
A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi
Oxana [17]

Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A

Therefore, for a  60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

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2 years ago
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A bicyclist passing through a city accelerates after he passes the sign post marking the city limits at x=0. His acceleration is
Anni [7]

Answer:

A. 24 m, 14 m/s

B. 8.0 m

Explanation:

Given:

x₀ = 6.0 m

v₀ = 4.0 m/s

a = 5.0 m/s²

t = 2.0 s

A. Find: x and v

x = x₀ + v₀ t + ½ at²

x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²

x = 24 m

v = at + v₀

v = (5.0 m/s²) (2.0 s) + (4.0 m/s)

v = 14 m/s

B. Find x when v = 6.0 m/s.

v² = v₀² + 2a (x − x₀)

(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)

x = 8.0 m

8 0
2 years ago
An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.
amm1812

Answer:

a) T ’= 0.999 s ,  b)  t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

        w = √g / L

The angular velocity, frequency and period are related

        w = 2π f = 2π / T

        2π / T = √ g / L

        T = 2π √ L / g

        L = T² g / 4π²

        L = 1² 9.8 / 4π²

        L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

       ΔL = α L ΔT

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       ΔL = 142.8 10⁻⁶ m

       Lf - L = -142. 8 10⁻⁶

       Lf = 142.8 10⁻⁶ + 0.248

       Lf = 0.2479 m

Let's calculate new period

      T ’= 2π √ L / g

      T ’= 2π √ (0.2479 / 9.8)

      T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

       t = 3600 0.999

       t = 3596.4 s

Therefore the clock is delayed almost 4 s

6 0
3 years ago
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