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Elan Coil [88]
3 years ago
14

If an 800.0 Kg sports car slows to 8.0 m/s in the forward direction to check out an accident scene and a 1200.0 kg pick-up truck

behind them continues traveling at 15.0 m/s forward, with what velocity will the two move if they lock bumpers after a rear-end collision
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

<h2><u>12.2m/s</u></h2>

Explanation:

Step one:

given data

mass m1=800kg

velocity v1=8m/s

mass m2= 1200kg

velocity v2=15m/s

Note: the two bodies move with the same velocity after collision, hence the common velocity is v=?

Step two:

the expression for the total momentum is

m1v1+m2v2=v(m1+m2)

substituting we have

800*8+1200*15=v(800+1200)

6400+18000=v(2000)

24400=2000v

divide both sides by 2000 to find v

v=24400/2000

v=12.2m/s

<u>Their common velocity is 12.2m/s</u>

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Answer:

wavelength \lambda = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = \frac{D\lambda}{d}    ....................1

\lambda is Wavelength of light and  D is Distance between screen and slit and d is slit width

so put here value and we get

\lambda = \frac{2.96*0.325*10^{-6}}{2.20}    

\lambda = 437.27 × 10^{-9} m

wavelength \lambda = 437.27 nm

4 0
3 years ago
Which of the following energy resources does not produce greenhouse
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Explanation:

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7 0
3 years ago
A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in:
Stells [14]

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in: 2 seconds

Answer:

2.4 m

Explanation:

From the question above,

Applying equation of motion,

s = ut+at²/2....................... Equation 1

Where t = time, u = initial velocity, a = acceleration,  s = distance.

make s the subject of the equation,

Given: a = 1.8 m/s²,  t = 2 seconds, u = 0 m/s (from rest)

Substitute these value into equation 1

s = 0(2)+1.8(2²)/2

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s = 2.4 m.

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4 0
3 years ago
When do vernal equinoxes usually occur? autumnal?
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7 0
3 years ago
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
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