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2 years ago

A student releases a marble from the top of a ramp. The marble increases

Physics

1 answer:

Fed [463]2 years ago

7 0

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at

Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)

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The answer is (a) because movement is acceleration

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2 years ago

You’ve made the hypothesis that the stepper the slope , the faster a ball will be rolling when it reaches the bottom .

BigorU [14]

Answer:

B) how steep the slope is

Explanation:

Because you have to know how is the influence of the steep of the slope in the time that a ball reaches the bottom. The steep of the slope is the variable that you would have to change in an experiment.

I hope this is useful for you

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4 0

3 years ago

.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric

maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0

3 years ago

How much ice (at 0Â°C) must be added to 1.90 kg of water at 79 Â°C so as to end up with all liquid at 8 Â°C? (ci = 2000 J/(kg.Â°

muminat

m = mass of the ice added = ?

M = mass of water = 1.90 kg

$c_{w}$ = specific heat of the water = 4186 J/(kg ⁰C)

$c_{i}$ = specific heat of the ice = 2000 J/(kg ⁰C)

$L_{f}$ = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

$T_{ii}$ = initial temperature of ice = 0 ⁰C

$T_{wi}$ = initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m $c_{w}$ (T - $T_{ii}$ ) + m $L_{f}$ = M $c_{w}$ ( $T_{wi}$ - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

4 0

3 years ago

State ohm’s law Q.34.7

defon

Ohm's law states that: $V=RI$

Explanation:

Ohm's law states that in a conductor, the potential difference across the conductor is directly proportional to the current flowing through it. Mathematically,

$V \propto I$

where

V is the potential difference

I is the current

The constant of proportionality is called resistance (R), and it gives a measure of "how much the conductor opposes" to the flow of current. Therefore Ohm's law can be rewritten as

$V=RI$

where R is the resistance. By rewriting the equation as

$I=\frac{V}{R}$

we see that the larger the resistance, the lower the current in the conductor.

Learn more about Ohm's law:

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4 0

3 years ago