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lara [203]
3 years ago
10

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea

st to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.42 m west and 6.37 m south of the impact point.How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Physics
1 answer:
skad [1K]3 years ago
4 0

Answer:

v = 11.1m/s

Vx = 7.19m/s

Vy = 8.45m/s

V(sedan) = 30.4m/s

V(suv) = 25.9m/s

Explanation:

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An orange light (f = 5.2 * 10'4Hz) is
KonstantinChe [14]

Answer:

2.145×10^-10 V or 0.2145nV

Explanation:

From hf=eV

h= Plank's constant = 6.6×10^-34JS

f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

Therefore the voltage created is 2.145×10^-10 V or 0.2145nV

7 0
3 years ago
We use coal, crude oil, and natural gas to generate electricity in _____________.
tresset_1 [31]

Answer:

<u>thermal power stations</u>

Explanation:

these resources are burned to produce the electricity.

5 0
2 years ago
Read 2 more answers
What is the impulse of a 3kg object accelerating from rest to 12m/s?
ale4655 [162]
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).

The answer is 36 kg m/s
6 0
3 years ago
As speed (velocity) increases, potential energy increases true or false
Lyrx [107]

Answer:

true

Explanation:

3 0
2 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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