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lions [1.4K]
3 years ago
12

Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car

b has a mass of 1,500kg. after the impact, the cars stick together and slide away at a speed of 7.1m/s. how fast was car a going when it hit car b?
Physics
1 answer:
frez [133]3 years ago
7 0

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

<span>vA = 13.02 m/s</span>

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Vector A has magnitude of 15.0 m/s and is 75° counter-clockwise up from the x-axis. What are the x- and y-components of the vect
LiRa [457]

Answer:

x component 3.88 y- component 14.488

Explanation:

We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure

Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )

y- component will be 15 sin 75°=14.488

For verification the resultant of x and y component should be equal to 15

So resultant =\sqrt{14.488^2+3.88^2}=15

3 0
3 years ago
Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?
musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

8 0
3 years ago
During which stage of sleep does most dreaming occur
kumpel [21]
REM, it is the deepest sleep and will send you deep within the mind
5 0
3 years ago
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0
3 years ago
Please answer above in the picture thanks​
Vera_Pavlovna [14]

Answer:1.14.102 on the side of 10

Explanation:

6 0
3 years ago
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