10. The force that causes damage to objects during a collision is

A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00

Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be $m_1$ and the 12.0 kg mass be $m_2$ . When they collide and stick, they have a combined mass of $m_1+m_2$ .

Momentum is given by $p=mv$ . Set up the following equation:

$\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f$ , where $v_f$ is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

$10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}$

4 0

2 years ago

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

5 0

4 years ago

The "Garbage Project" at the University of Arizona reports that the amount of paper discarded by households per week is normally

Lorico [155]

Answer:

54%

Explanation:

We are given that

S.D=4.2 lb

Mean= $\mu=9.4 lb$

We have to find the percentage of household throw out at least 9 lb of paper a week.

Normal distribution formula :

$P(X\geq a)=P(z\geq \frac{a-\mu}{\sigma})$

We have a=9

$P(X\geq 9)=P(z\geq \frac{9-9.4}{4.2})=P(z\geq -0.1)$

$P(X\geq 9)=1-P(z$

$P(X\geq 9)=1-0.4602=0.5398\times 100=54$ %

Hence, the percentage of household throw out at least of paper a week=54%

7 0

3 years ago

What is defined as a special kind of systematic observation, used by correlational researchers,

erica [24]

Longitudinal design is defined as a special kind of systematic observation, implemented by correlational researchers, which involves the tasks of obtaining measures of variables of interest in multiple waves over time.

<u>Explanation: </u>

Longitudinal design is one of the known systematic observation pattern followed by the correlational researchers in which a subject of experiment is observed for a significantly long span.

The three types of longitudinal design are: panel study, cohort study and retrospective study. The complete design involves the multiple set of observations over a variable or subject. Longitudinal design is often used in clinical psychology to study rapid fluctuations in the behaviour.

7 0

4 years ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago