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3 years ago

7

Pace left home at 8 AM to spend the day at an amusement park. He arrived at the park, which was 150 km from his house, at 10 AM.

Pace explained that his speed was 38 km/h, which was a/an (accelerated / directional / instantaneous / average) speed, but the fastest speed he went was 60 km/h, which was a/an (directional / instantaneous / accelerated / average) speed.

1 answer:

arsen [322]3 years ago

3 0

Answer:

average speed, accelerated speed

Explanation:

Pace was first moving at 38 km/h which was his average speed but the fastest speed he went was 60 km/h so he accelerated.

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A particle travels along a straight line with a velocity v = (12 - 3t) m/s, where t is in seconds. When t = 1 s, the particle is

Simora [160]

Answer:

(1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

Explanation:

Given that,

Velocity = (12-3t^2) m/s

When t = 1 s, the particle is located 10 m to the left of the origin.

We need to calculate the acceleration at t = 4 sec

Using formula of acceleration

$a=\dfrac{dv}{dt}$

Put the value of v

$a=\dfrac{d}{dt}(12-3t^2)$

$a=-6t$

Put the value of t

$a=-6\times4$

$a=-24\ m/s^2$

The displacement from t = 0,t = 10 s, and the distance the particle travels during this time period.

We need to calculate the distance

Using formula of distance

$ds=v\ dt$

$\int_{-10}^{s}=\int_{1}^{t}{v}dt$

Put the value of v

$\int_{-10}^{s}=\int_{1}^{t}{ (12-3t^2)}dt$

$s+10=12t-t^3-11$

$s=12t-t^3-21$

At t = 0,

$s_{t=0}=-21$

At t = 10,

$s_{t=10}=12\times10-10^3-21$

$s_{t=10}=-901$

The displacement is

$\Delta s=-901-(-21)$

$\Delta s=-880\ m$

The distance at t= 2 sec

$s_{t=2}=12\times2-2^3-21$

$s_{t=2}=-5$

The total distance will be,

$s_{T}=(21-5)+(901-5)$

$s_{T}=912\ m$

(2). We need to calculate the distance at 2 sec

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$s=27\times 2+\dfrac{1}{2}\times6\times(2)^3$

$s=78\ m$

Hence, (1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

3 0

3 years ago

At T = 0C and p = 1000mb, 1 g of dry air receives an amount of heat during an isochoric process. It is then observed that its p

Rufina [12.5K]

Answer:

ΔT = 13.65° C

ΔQ = 13.7 J

Explanation:

First we will find the final temperature of air by using equation of state:

P₁V₁/T₁ = P₂V₂/T₂

For Isochoric Process, V₁ = V₂

Therefore,

P₁/T₁ = P₂/T₂

T₂ = P₂T₁/P₁

where,

T₂ = Final Temperature = ?

P₂ = Final Pressure = 1050 mb

P₁ = Initial Temperature = 1000 mb

T₁ = Initial Temperature = 0°C = 273 k

Therefore,

T₂ = (1050 mb)(273 K)/(1000 mb)

T₂ = 286.65 K

Change in Temperature = ΔT = T₂ - T₁

ΔT = 286.65 K - 273 K

<u>ΔT = 13.65° C</u>

<u></u>

The first law of thermodynamics can be written as:

ΔQ = ΔU + W

where,

ΔQ = heat absorbed

ΔU = change in internal energy = mCΔT

W = Work Done = 0 (in case of isochoric process)

Therefore.

ΔQ = mCΔT

where,

m = mass of air = 1 g = 1 x 10⁻³ kg

C = specific heat of dry air = 1003.5 J/kg.°C

Therefore,

ΔQ = (1 x 10⁻³ kg)(1003.5 J/kg.°C)(13.65°C)

<u>ΔQ = 13.7 J</u>

7 0

3 years ago

What will happen to the period of a ball-on-a-spring if the ball is replaced by a smaller one with half the mass

Murrr4er [49]

When the ball on a spring is replaced by a smaller ball with half the mass : The period decrease by 29%

<h3>Calculate the Period </h3>

We will apply the formula below to calculate period

T = $2\pi \sqrt{\frac{m}{k } }$

where :

T₁ / T₂ = $\sqrt{\frac{m_{1} }{m_{2} } }$ where m₂ = m₁ / 2

Therefore :

T₂ = $\frac{T_{1} }{\sqrt{2} }$ = 0.71 T

Hence we can conclude that the period will decrease by 29% ( i.e 100 - 71 ).

Learn more about Period calculation : brainly.com/question/10728818

4 0

2 years ago

Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the hori

lys-0071 [83]

Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - $\frac{32.2 ft/s^{2} *3.92 s^{2} }{2}$

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32

6 0

3 years ago

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

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

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0

4 years ago