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3 years ago

14

A speeding car collides with a wall (attached to Earth). Consider the car-Earth system to be isolated. A loud sound is produced

in the collision, and the car is deformed. What can you say about the momentum of the car-Earth system in this case?

2 answers:

AleksAgata [21]3 years ago

8 0

That the total momentum of the system is conserved

allsm [11]3 years ago

7 0

Answer:

Momentum of car and Earth as a system will remain conserved

Explanation:

As we know that momentum conservation is applicable when system has no external force on it

So here we will say

$F = \frac{\Delta P}{\Delta t}$

now we know that

if we will have F = 0

$0 = \frac{\Delta P}{\Delta t}$

so here we can say

$\Delta P = 0$

now since Earth and car as a whole system is considered then there is no external force on that system.

So here momentum as a whole will remain conserved.

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You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

John pushes a box with a constant force as shown in the graph below.

andrew11 [14]

From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.

4 0

3 years ago

Which question cannot be answered through making measurements?

Bond [772]

A. is the answer for this question

5 0

4 years ago

A skydiver of mass 80kg jumps from a slow moving aircraft and reach a terminal speed of 50 m/s. What's her acceleration when her

miskamm [114]

Answer:

a = g = 9.81[m/s^2]

Explanation:

This problem can be solve using the second law of Newton.

We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.

m*g = m*a

where:

g = gravity = 9.81[m/s^2]

a = acceleration [m/s^2]

Note: If the skydiver will be under air resistance forces his acceleration will be different.

7 0

3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago