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mezya [45]
3 years ago
12

A boat heads north at 30 mph while a steamer heads east at 40 mph at the same time. When is the distance between them 50 mi? Ans

wer:A boat heads north at 30 mph while a steamer heads east at 40 mph at the same time. When is the distance between them 50 mi? Answer:
Physics
1 answer:
Reil [10]3 years ago
7 0

Answer:

1 hour

Explanation:

Speed of the first boat = 30 mph

Speed of the second = 40 mph

The boats will cover different distances but the time taken will be the same.

Time taken by the boats = t

Distance = Speed × Time

Distance covered by the first boat = 30t

Distance covered by the second boat = 40t

Distance between the boats = 50 mi

From the Pythagoras theorem

\sqrt{(30t)^2+(40t)^2}=50\\\Rightarrow \sqrt{2500t^2}=50\\\Rightarrow 50t=50\\\Rightarrow t=\frac{50}{50}=1\ hour

Time taken by the boats when they are 50 mi away is 1 hour

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Speed and direction are the two scientific componets that make up velocity.

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2 years ago
From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h
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Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

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v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

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3 0
3 years ago
Read 2 more answers
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
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g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin
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Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as mgh=0.5mv^{2} since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then v=\sqrt {2gh} and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s

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Marine magnetic anomalies result from seafloor spreading in conjunction with
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