Triangle FGH is pictured below.

a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance

8_murik_8 [283]

The car's (average) acceleration would be

$a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}$

The car's position over time would be given by

$x=v_0t+\dfrac12at^2$

so that after 2.4 seconds, the car will have traveled a distance of

$x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2$

$\implies x=77.5\,\mathrm m$

7 0

3 years ago

Read 2 more answers

A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. If it falls the second half

Hitman42 [59]

Answer:

$t2/t1 = \sqrt{2} - 1$

Explanation:

The expression for the second law of motion is given below:

h = $ut + 0.5at^2$

<u>For first half distance</u>

Object is initially at rest, so its initial speed u = 0

Object falls at half the distance, so h = h/2 where t = t1

Hence, we have

$h/2 = at1^2/2 - equation 1$

<u>For second half distance: </u>

Similarly,

$h = a(t1 + t2)^2/2 - equation 2$ where t = t1 + t2 and u= 0

Using equation 2 by equation 1

we obtain $2 = (t1 + t2)^2/t1^2$

Hence $t2/t1 + 1 = \sqrt{2}$

Hence $t2/t1 = \sqrt{2} - 1$

5 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0

Grace [21]

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0

3 years ago

How many Joules of potential

s2008m [1.1K]

Answer:

40 j, 80j.

Explanation:

P.E= mgh. G=10 m/s².

For 4m, P.E=1*10*4=40 joules.

For 8m, P.E=1*10*8=80 joules.

4 0

4 years ago

For a railgun, determine the current required to accelerate a 1000kg space vehicle to earths escape velocity on 10km of track wi

Troyanec [42]

Answer:

the current is 15.68 × $10^{5}$ A

Explanation:

mass = 1000 kg

magnetic fied = 2T

rail separation = 2m

escape velocity is 11.2km/s = 11.2 × $10^{3}$ m/s

distance = 10 km = $10^{4}$ m

to find out

determine the current

solution

we know force F = I×L×B

here I is current and L is rail separation and B is magnetic field

so F = I ×2×2 = 4 I

so

acceleration is a = $\frac{F}{mass}$

a = $\frac{4I}{1000}$ m/s²

so equation of motion

v²-u² = 2 a S

here u is initial velocity and S is distance and a is acceleration and v is final velocity

11.2 × $10^{3}$ - 0 = 2× $\frac{4I}{1000}$ × $10^{4}$

solve we get I

I = 15.68 × $10^{5}$ A

so the current is 15.68 × $10^{5}$ A

4 0

3 years ago