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KonstantinChe [14]
4 years ago
7

A car with mass mc = 1490 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck o

f mass mt = 1650 kg traveling south at vt = 6.4 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5. write equation of the velocity of the system after collision.
Physics
1 answer:
icang [17]4 years ago
3 0

Answer:

v= - 4.507 i - 2.363 j

Explanation:

 Given that

mc= 1490 kg

vc= 9.5 m/s ( - i)

mt=  1650 kg

vt = 6.4 m/s ( -j)

There is any external force so linear momentum will remain conserve.

Lets take final speed is v.

mc .vc + mt . vt = ( mc+mt) v

1490 x 9.5 ( - i) + 1650 x 6.4 ( -j) = ( 1490+1650) v

14,155 ( -i) + 10,560 ( - j) = 3140 v

v= - 4.507 i - 2.363 j

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A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling
IrinaVladis [17]

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

-u as it is moving in opposite direction.

mc = 1200 kg, uc = 31.1 m/s

ms = 2830 kg, us = - u = ?

Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

4 0
4 years ago
The sun is 1.50x10^11 m from earth. How long does it take the suns light to reach earth? How long
galina1969 [7]

Answer:

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3 0
3 years ago
An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground t seconds after the fall is ​s(t)
Paul [167]

Answer: v= 160ft/s

a=32ft/s^2 constant

Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction

7 0
3 years ago
Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm.In this confi
Sladkaya [172]

Answer:

The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

F = k\dfrac{q_1 \times q_2 }{r^2}

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.

8 0
4 years ago
Which of the following has the lowest U value?
jarptica [38.1K]

Answer:

c. expanded polyurethane

Explanation:

Thermal performance of a building fabric is measured in terms of heat loss and is expressed as U-value or R-value. U-value is the rate of heat transferred through a structure divided by the difference in temperature across the structure with a unit of measurement of W/m²K.You can calculate the U-value of a by getting the reciprocal of the sum of thermal resistances , R, making the building material.

If you have the value of R, then U=1/R

Material                         size            R                      U

plywood                          1"              1.25                0.8

Poured concrete            2"              0.99               1.010

Expanded polyurethane  1"            6.5                   0.1538

Asbestos shingles            1"             0.03                33.33

The material with lowest U-value is expanded polyurethane

4 0
3 years ago
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