An object is stopped at point ‘a’, and travels to point ‘c’ in 8s. What was its acceleration?

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

ale4655 [162]

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

3 0

2 years ago

A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 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

Based on what you have just read, what are the advantages of storing music digitally compared to using an analog system?

melamori03 [73]

Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.

7 0

3 years ago

Read 2 more answers

When an alkali metal is involved in a chemical reaction with a nonmetal, what charge will its atoms most likely form? 1- 2- 2+ 1

seraphim [82]

Answer:

+1 ion

Explanation:

Alkali metals are metals that are found in Group I of the periodic table. Their electronic configuration is such that their valence shell in grounds state is always holding only one electron which they always lose when reacting with non-metals. A loss in an electron makes the atom electrically imbalanced and hence becoming a +1 ion.

7 0

3 years ago