Ask question

Ask question

All categories

3 years ago

14

A car is driving at a velocity of 24 m/s.if it brakes can supply an acceleration of -5.0 m/s^2, how much time will be required t

o bring the car to a stop.

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer:The time required can be calculated by the following formula;

Explanation:

You might be interested in

Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 3 s. What is your acceleration?

lisov135 [29]

You asked the question twice I answered it on the last one

7 0

3 years ago

What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms

Alexus [3.1K]

Answer:

1.5F

Explanation:

Using

E= F/q

Where F= force

E= electric field

q=charge

F= Eq

So if qis tripled and E is halved we have

F= (E/2)3q

F= 1.5Eq=>> 1.5F

4 0

2 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

A person driving her car at 48 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow

trasher [3.6K]

Answer:

She must stop the car before interception, distance traveled 12.66 m

Explanation:

We will take all units to the SI system

Vo = 48Km / h (1000m / 1Km) (1h / 3600s) = 13.33 m / s

V2 = 70 Km / h = 19.44 m / s

We calculate the distance traveled before stopping

X = Vo t + ½ to t²

Time is what it takes traffic light to turn red is t = 2.0 s

X = 13.33 2 + 1.2 (-7) 2²

X = 12.66 m

It stops car before reaching the traffic light turning to red

Let's analyze what happens if you accelerate, let's calculate the acceleration of the vehicle

V2 = Vo + a t2

a = (V2-Vo) / t2

a = (19.44-13.33) /6.6

a = 0.926 m / s2

This is the acceleration to try to pass the interception, now let's calculate the distance it travels in the time the traffic light changes from yellow to red (t = 2.0 s)

X = Vo t + ½ to t²

X = 13.33 2 + ½ 0.926 2²

X = 28.58 m

Since the vehicle was 30 m away, the interception does not happen

4 0

3 years ago

Fe₂O3<br> + co<br> →<br> Fe3O4 + CO₂

Goryan [66]

Explanation:

Fe₂O₃ + CO → Fe₃O₄ + CO₂

Balancing the equation above, we can derive simple mathematical equations that are very easy to solve.

aFe₂O₃ + bCO → cFe₃O₄ + dCO₂

a,b,c and d are the coefficients needed to balance the equation above;

Conserving Fe; 2a = 3c

O: 3a + b = 4c + 2d

C: b = d

let a = 1;

c = $\frac{2}{3}$

Since b = d

3a + d = 4c + 2d

3a = 4c + 2d - d

3a = 4c + d

a = 1, c = $\frac{2}{3}$

3 = 4 x $\frac{2}{3}$ + d

d = $\frac{1}{3}$

b = $\frac{1}{3}$

multiplying a, b, c and d by 3:

a = 3 b = 1 c = 2 and d = 1

3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

Learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

6 0

3 years ago