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2 years ago

Please the organisms in the order of transmission for Lyme disease. Begin with the organism that originally carries the bacteria

Physics

1 answer:

Aleonysh [2.5K]2 years ago

5 0

White footed mouse, than human.

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A man walks 2.6 miles east, then turns and walks 6.7 miles north. What is his resultant

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Answer:

Yes cause he walks 6.7 miles

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2 years ago

What is The distance traveled by a vehicle in 12 minutes,if it’s speed is 35km/h

Delvig [45]

Answer: 7 km

Explanation: 12 minutes is 1/5 of an hour. 35/5=7

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3 years ago

In the diagram of the conical pendulum below, which number (1, 2, 3, or 4) represents the angle θ used in the conical pendulum e

Mama L [17]

Step 1: Look in your book or online for the conical pendulum equation.

Step 2: Look at the drawing and see which angle is involved in the equation.

Answer: It's Angle #2 in your drawing.

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2 years ago

If a car accelerates at a uniform 4.0 m/s, how long will it take to reach a speed of 36.0 m/s,

11111nata11111 [884]

Answer:

Explanation:

v=u+at

36=0+4t

t=36-0/4

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2 years ago

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

Case for the rock $m = 5.0 Kg$ :

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

Case for the pebble $m = 3.0 Kg$ :

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

2 years ago

Read 2 more answers