What is a concern about recovered memories?

Two cars collide at an intersection. Car A , with a mass of 2000kg , is going from west to east, while car B , of mass 1400kg ,

ahrayia [7]

Complete Question:

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

Answer:

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

7 0

3 years ago

Step by step process on how to solve this problem

rusak2 [61]

Step-1: Remember or look up the formula for the force of gravity between two objects. F = G · m₁ · m₂ / R²

Step-2: Remember or look up the value of G. G = 6.67 x 10⁻¹¹ m³·kg/s²

Step-3: Write the numbers you know into the formula.

(1.989 x 10²⁰ Newtons) =

(6.67 x 10⁻¹¹ mtr³·kg/s²) · (5.9742 x 10²⁴ kg) · (moon mass) / (3.84 x 10⁸ mtr)²

Step-4: Sit back, relax, take your time, look this mess over, carefully, end-to-end, and decide how to solve it for (moon mass) .

Step-5: Divide each side by (6.67x10⁻¹¹mtr³·kg/s²)·(5.9742x10²⁴kg)/(3.84x10⁸mtr)²:

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

Step-6: Crunch the numbers. Be careful to KEEP all the units as you go along. When you're done, the units of your answer will be the first instant indication if you made a mistake. You're looking for the MASS of the moon. If the answer doesn't have units of 'kg', then that'll be an immediate red flag, telling you that there's been a mistake somewhere.

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

Collect the numbers, and collect the units:

Moon mass = (1.989x10²⁰ · (3.84x10⁸)² / (6.67x10⁻¹¹ · 5.9742x10²⁴)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = (1.989 · 3.84² x 10³⁶) / (6.67 · 5.9742 x 10¹³)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = 0.736 x 10²³ kg

Moon mass = 7.36 x 10²² kg

Step 7: Look it up in a book or online. See if you're anywhere close.

When I search "moon mass" on Floogle, the first hit says

" 7.348 x 10²² kg " .

yay ! I'm satisfied.

6 0

3 years ago

Estimate the average power output of the sun, given that about 1350 w/m2 reaches the upper atmosphere of the earth.

fenix001 [56]

the average power output of the sun is 3.8 × $10^{26}$ W given that about 1350 w/m2 reaches the upper atmosphere of the earth.

As we know intensity falling over a surface is equal to Energy falling per unit area per unit volume.

I = $\frac{Energy}{area*time}$

Since Power is equal to energy/time

I = power/area

or

Power = Intensity × area

Now, Distance from Sun to Earth , r = 149.6 × $10^{9}$ m

So, Surface area of the sphere of radius is:

A = 4 π $r^{2}$

= 4 × 3.14 × (149.6 × $10^{9}$ )²

= 2.81 × $10^{23}$ m²

Thus Average Power output of the sun will be:

P = 1350 × 2.81 × $10^{23}$

= 3.7935 × $10^{26}$

P = 3.8 × $10^{26}$ W

So the average power output of the sun is 3.8 × $10^{26}$ W.

If you need to learn more about about the average power of a resistor, click here.

brainly.com/question/12972958

#SPJ4

7 0

1 year ago

Read 2 more answers

In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to

Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

$\beta = \frac{\lambda D}{d}$

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

4 0

3 years ago

Which of these statements best explains why models are used to study atoms? Atoms of all materials are not the same. Atoms canno

Fed [463]

Answer by YourHope:

Hi! :)

Which of these statements best explains why models are used to study atoms?

Atoms are too small to be examined in detail without a microscope!

Models help scientists to study atoms because atoms are extremely small and can't even be seen!

:)

3 0

3 years ago