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2 years ago

A crazy dog runs at a constant speed of 19.85 mi/hr for 6.09 min. How far does the dog travel during this time period?

Physics

1 answer:

Phantasy [73]2 years ago

8 0

Answer:

Distance, d = 3242.19 meters

Explanation:

It is given that,

Speed of the dog, v = 19.85 mi/hr

Since, 1 mph = 0.44704 m/s

v = 19.85 mi/hr = 8.873 m/s

Time taken by the dog, t = 6.09 min = 365.4 sec

Let the distance covered by the dog during this time period is d. It can be calculated by the speed of the dog multiplied by the time taken as :

$d=v\times t$

$d=8.873\times 365.4$

d = 3242.19 meters

So, the the dog travel during this time period is 3242.19 meters. Hence, this is the required solution.

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What you do is, multiply 16.0 and 12.4 together. then multiply that by 40a

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3 years ago

the presence of which magnetic feature best explains why a magnet can act a distance on other magnets or on objects containing c

katen-ka-za [31]

Magnetic fields

Explanation:

The presence of magnetic fields best explains why a magnet can act a distance on other magnets or on objects containing certain metals.

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3 years ago

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

jenyasd209 [6]

Answer:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

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3 years ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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2 years ago

How can you define a solution to an equation?

sleet_krkn [62]

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3 years ago