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3 years ago

12

Complete this sentence. Average speed is worked out from dividing _ by

2 answers:

mestny [16]3 years ago

4 0

Average speed is worked out from dividing the total distance by the total time.

Dmitry [639]3 years ago

3 0

Average speed is worked out from dividing distance by time.

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A missile is fired from a jet flying horizontally at Mach 1 (1100 ft/s). The missile has a horizontal acceleration of 1000 ft/s2

nadezda [96]

Answer: 11,100 ft/s^2

1) Constant acceleration=> uniformly accelerated motion.

2) Formula for uniformly accelerated motion:

Vf = Vo + at

3) Data:

Vo = 1,100 ft/s
a = 1,000 ft/s^2
t = 10.0 s

4) Solution:

Vf = 1,100 ft/s + 1,000 ft/s^2 * 10.0 s = 1,100 ft/s + 10,000 ft/s

Vf = 11,100 ft/s

6 0

3 years ago

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The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

4 years ago

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A scientist is measuring various properties of a sound wave. She measures the value 340 m/s. Which of the following wave charact

n200080 [17]

It’s either. A or D

8 0

2 years ago

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1. If a model train car with a momentum of 12 kg•m/s collides with a 2 kg model train car that is not moving, what is the total

wariber [46]

Answer:

$P_{f} = 12 \ kg.m/s$

Explanation:

Given data:

Momentum of moving model train, $P_{1} = 12 \ kg.m/s$

Mass of the stationary model train, $m = 2 \ kg$

Initial speed of the stationary model train, $v = 0$

Assume there is no external force is acting on the given train system.

In this case, the total linear momentum of the trains would be conserved.

Let the final linear momentum of the trains be $P_{f}.$

Thus,

$P_{i} = P_{f}$

$P_{1} + P_{2} = P_{f}$

$P_{1} + mv = P_{f}$

$12 + 2 \times 0 = P_{f}$

$\Rightarrow \ P_{f} = 12 \ kg.m/s.$

7 0

3 years ago

A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is

Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2$

The acceleration of the craft should be 1.02234 m/s²

$F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N$

Weight of the craft

$W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N$

Thrust

$F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N$

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0

4 years ago