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Dmitriy789 [7]
3 years ago
10

Hey! Can someone help with this question? Thx :)

Physics
1 answer:
Schach [20]3 years ago
4 0
<span>A. Chemical energy to chemical energy</span>
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A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together
STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0
3 years ago
A seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False
iris [78.8K]
The answer is A, True.
7 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
mixer [17]

Answer:

fffffffffffffffffffffffffffff

5 0
3 years ago
Read 2 more answers
What is the frequency of a sound wave directly related to?
Mama L [17]
The frequency of a sound wave is the number of cycles of a sound wave per second or Hertz (Hz). The frequency can be calculated by dividing wavelength by time (Figure 1.3). A small wavelength will yield a higher frequency, whereas a larger wavelength will yield a smaller frequency.
8 0
3 years ago
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