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2 years ago

15

Radon is the heaviest naturally radioactive ________________ gas.

1 answer:

MariettaO [177]2 years ago

7 0

Random is the heaviest naturally radioactive NOBLE gas

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The balls in the image above have different masses and speeds. Rank them in terms of momentum, from least to greatest.

N76 [4]

The momentum of an object is equivalent to the product of the object's mass and velocity. Computing the momentum for each ball:
A- 15 * 0.7 = 10.5
B- 5.5 * 1.2 = 6.6
C- 5.0 * 2.5 = 12.5
D- 1.5 * 5.0 = 7.5

Therefore, ball C has the greatest momentum.

3 0

2 years ago

Read 3 more answers

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0

2 years ago

If the frequency of a periodic wave is doubled, the period of the wave will be

Lunna [17]

The period of the wave would be halved

5 0

2 years ago

Using the picture above, which ball has the greatest potential energy?

weqwewe [10]

Ball 4 because the higher the elevation is the greater the potential energy it has

4 0

2 years ago

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

2 years ago