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3 years ago

A traveler covers a distance of 413 miles in a time of 7 hours. What is the average speed for his trip?

Physics

1 answer:

Nina [5.8K]3 years ago

7 0

Answer:

59/mph.

Explanation:

We just have to divide the number of miles, by the amount of time it took. 413 miles / 7 hours = 59 miles per hour.

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A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second

Drupady [299]

<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation. F = Ma F is the net force M is the total mass of the system a is the acceleration a = F / M a = (mb)(g) / (ma + mb) a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg) a = 58.8 N / 20 kg a = 2.94 m/s^2 The magnitude of the acceleration of the system is 2.94 m/s^2</span>

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3 years ago

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

7 0

3 years ago

A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km