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3 years ago

A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration

Physics

1 answer:

LuckyWell [14K]3 years ago

6 0

<u>Answer</u>

The acceleration is

$a=-2.5ms^{-2}$ to the nearest tenth

<u>Explanation</u>

Since the car was travelling at $37ms^{-1}$ before it starts to decelerate, the initial velocity is

$u=37ms^{-1}$ .

The final velocity is $v=0ms^{-1}$ , because the car came to a stop.

The time taken is $t=15s$ .

Using the Newton's equation of linear motion,

$v=u +at$ , we find the acceleration by substituting the known values.

This implies that,

$0=37 +a(15)$

This gives us,

$0-37=15a$

$\Rightarrow -37=15a$

We divide both sides by 15 to get,

$a=-\frac{37}{15}ms^{-2}$

$a=-2.46667ms^{-2}$

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2 years ago

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A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

Norma-Jean [14]

Answer:

Now e is due to the ring at a

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

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8 0

3 years ago

A certain resistance thermometer read 14.5 ohms in pure melting ice and 18.5 ohms in steam at standard atmospheric pressure what

Vadim26 [7]

The resistance of the thermometer at room temperature is 15.04 ohms.

<h3 />

<h3>What is a resistance thermometer?</h3>

A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.

To calculate the resistance of the thermometer at room temperature, we use the formula below.

Formula:

100/27 = 2/(x-14.5)..............Eqquation 1

Where:

x = Resistance of the thermometer at room temperature

Make x the subject of the equation

x = [(27×2)/100]+14.5
x = (54/100)+14.5
x = 0.54+14.5
x = 15.04 ohms.

Hence, The resistance of the thermometer at room temperature is 15.04 ohms.

Learn more about thermometers here: brainly.com/question/1531442

3 0

1 year ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

2 years ago

A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

$(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]$

7 0

2 years ago