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OverLord2011 [107]

3 years ago

6

Which contributes most to our average dose of background radiation?

1 answer:

Anuta_ua [19.1K]3 years ago

6 0

C) Natural and Artificial sources

You might be interested in

When you look ahead while driving, it is best to:

vodka [1.7K]

I think its [B]
Personally i would say [B] only because If you are looking beyond the car in front of you..... then what if the car in front of you throws on breaks... you would hit them in the butt because you weren't paying attention to the car.
And majority of the time if your looking in the lanes beside you then you are most likely trying to get in that lane.

4 0

3 years ago

Select three possible applications of a capacitor. Select all that apply.

Wittaler [7]

Answer:

I'm pretty sure it's all of them i'm not completely sure though hope it helps anyways! :)

7 0

3 years ago

Pls help<br> What graph indicates no reference to direction of motion, and explain why.

Komok [63]

Answer:

graph A

Explanation:

the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)

but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

4 0

3 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

<u>Where:</u>

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

3 years ago

3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the

const2013 [10]

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car = 7.65 x 10³N

Unknown:

Acceleration of the car = ?

Solution:

From Newton's second law of motion:

Force = mass x acceleration

Acceleration = $\frac{Force }{mass}$ = $\frac{7.65 x 10^{3} }{2.5 x 10^{3} }$ = 3.06m/s² to the east

6 0

3 years ago