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2 years ago

21. Compare and contrast beta particles to the electrons that surround neutral atoms.

1 answer:

6 0

Beta particles come from the nucleus. Electrons are found around the nucleus.
Beta particles normally travel very fast out of a nucleus in a straight line. Electrons normally orbit the nucleus of an atom.

They both have the same mass and the same charge.

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The soccer game begins with a ______ ______ in the center of the field

blagie [28]

It begins with a kick-off

6 0

3 years ago

A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu

bija089 [108]

Answer:

J=36221 Kg.m/s

Explanation:

Impulse-Momentum Theorem

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

$p=mv$

The initial and final momentums are, respectively

$p_1=mv_1,\ p_2=mv_2$

The change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It is numerically equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

We are given

$m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s$

The impulse the car experiences during that time is

$J=2498(2.6-17.1)=2498(-14.5)$

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

3 years ago

. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe

julia-pushkina [17]

By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P₂*50

P₂*50= 300*75

P₂ = 300*75/50 = 450

P₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.

3 0

2 years ago

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Define the term “force”.

Aleonysh [2.5K]

Energy that is applied to an object.

--TheOneandOnly003

3 0

3 years ago

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