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Temka [501]
3 years ago
12

A compact car with a mass of 1,000 kg, traveling at 15 m/s, runs into and sticks to - an at-rest SUV of 2,200 kg. What is the fi

nal speed of the two cars? A. 4.7 m/s B. 67 m/s C. 2.2 m/s D. 147 m/s
Physics
1 answer:
zepelin [54]3 years ago
4 0
Roughly 4.7 m/s this would be a momentum argument as well
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A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?
VLD [36.1K]
The initial kinetic energy of the car is
E_1 =  \frac{1}{2}mv_1^2 =  1.9 \cdot 10^3 J

Then, the velocity of the car is decreased by half: v_2 =  \frac{v_1}{2}
so, the new kinetic energy is
E_2 =  \frac{1}{2}mv_2 ^2 =  \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
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5 0
4 years ago
What is the relationship between temperature and altitude in the stratosphere? (2 points).
stealth61 [152]
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8 0
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A child lifts a 5.0-newton toy to a height of 0.50 meters. How much work is done on the toy?
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6 0
4 years ago
Read 2 more answers
For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What
choli [55]

Answer

given,

Stress for plastic deformation =  267 MPa

modulus of elasticity = 115 GPa

cross sectional area = 377 mm²

a)    maximum load (in N) that may be applied to a specimen

= σ x A

= 267 x  10⁶ x  377 x 10⁻⁶

= 100659 N

b)   modulus of elasticity = stress/strain

     115 x 10⁹  =\dfrac{267 \times 10^6}{\dfrac{\Delta l}{L}}

        L = 127 mm

      115 x 10⁹  =\dfrac{267 \times 10^6}{\dfrac{\Delta l}{127}}

      \dfrac{\Delta l}{127}=\dfrac{267 \times 10^6}{115\times 10^9}

    Δ l =   0.295 mm

maximum length after the stretched = 127 mm + 0.295 mm

                                                            = 127.295 mm

4 0
3 years ago
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