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Vedmedyk [2.9K]
3 years ago
5

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv

e direction, what are John and Betty's speeds? A. -20 meters/second, 20 meters/second B. 20 meters/second, 20 meters/second C. -20 meters/second, -20 meters/second D. 0 meters/second, 40 meters/second
Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
The answer is definitely A
rewona [7]3 years ago
7 0

Answer:

B. 20 meters/second, 20 meters/second

Explanation:

Velocity of John = 20 meters/second north = +20 meters/second

Velocity of Betty =  20 meters/second south = -20 meters/second

If north is considered as positive then south will be negative.

So, the velocity of John will be +20 meters/second and velocity of Betty will be -20 meters/second.

But, the question is about their speeds and not velocity. The signs represent the directions. Speed does not show direction.

So, speed of John and Betty is 20 meters/second.

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A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?
EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by d = \frac { m } { v }.

Here in this problem, we are given a mass of 565 g which occupies a volume of 50 cm^3.

So plugging the data in the above formula to find the density:

Density = \frac { 565 } { 50 } = 11.3

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0
3 years ago
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
3 years ago
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

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3 years ago
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Leokris [45]
It is called a homologous chromosome meaning it carries the same gene
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3 years ago
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
3 years ago
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