Please help me with this Science Worksheet

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

-BARSIC- [3]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62 distance traveled after collision .. crumpling of truck

Part a

$V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s$

Part b

$vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s$

7 0

3 years ago

Read 2 more answers

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

I. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maxi

nalin [4]

Answer:

Negative

Explanation:

If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.

Hope this helped. :)

6 0

3 years ago

Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?

Bond [772]

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

7 0

3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,