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Taya2010 [7]
3 years ago
14

Please help! Drag and drop the words into the correct locations

Physics
2 answers:
Stells [14]3 years ago
7 0

Answer:

background radiation

uniform

13700000000 years ago

vaieri [72.5K]3 years ago
5 0

Answer:

Explanation:

1. bACk gROuND rADiATIon

2. nOT UnIFOrm

3. 17,300,000,000 yEarS agO

You might be interested in
State the Newton 2nd law of motion and also prove that F= ma<br>​
saw5 [17]

Explanation:

F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is defined as Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

...

3 0
3 years ago
What are the three places where ribisomes occur in a cell​
irina [24]

Answer: Rough endoplasmic reticulum, cytoplasm, inside mitochondria

Explanation:

4 0
3 years ago
Net force of 8.0 N acts on an 18 kg body for one minute. Determine the impulse due to the force.
boyakko [2]

Answer:

p = FΔt = 8.0 N(60 s) = 480 N•s

Explanation:

not asked for, but in that time a frictionless 18 kg mass on a horizontal surface will have change velocity by 480/18 = 26.7 m/s.

An impulse results in a change of momentum.

3 0
3 years ago
Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
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