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LUCKY_DIMON [66]

3 years ago

13

Earth is the only planet able to support what

2 answers:

saw5 [17]3 years ago

8 0

Life... obviously lol

exis [7]3 years ago

4 0

The answer is actually oxygen

You might be interested in

3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

inessss [21]

Answer:

The average angular acceleration is $\alpha =125.487 rad /s^2$

Explanation:

From the question we are told that

From the question we are told that

The length of the bat is $l = 0.85m$ \

The initial linear velocity is $u = 0 m/s$

The time is $t = 0.15s$

The velocity at t is $v = 16 m/s$

Generally average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_o}{t}$

Where $w_f$ is the finial angular velocity which is mathematically evaluated as

$w_f = \frac{v}{l}$

$w_f = \frac{16}{0.85}$

$= 18.823 rad/s$

and $w_o$ is the initial angular velocity which is zero since initial linear velocity is zero

So

$\alpha = \frac{18.823 - 0}{0.15}$

$\alpha =125.487 rad /s^2$

5 0

3 years ago

(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the

frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a. $C=8\mu F=8\times 10^{-6} F$

$1\mu =10^{-6}$

Q= $48\mu C=48\times 10^{-6} C$

We know that

$V=\frac{Q}{C}$

Using the formula

$V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V$

b. $Q=192\mu C=192\times 10^{-6} C$

$V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V$

8 0

3 years ago

An isotope is a version of the same element that differs in the composition of the A) atom. B) nucleus. C) particle. Eliminate D

crimeas [40]

The correct answer is B because isotopes have different numbers of neutrons, and neutrons are located in the nucleus

Hope this helps

5 0

3 years ago

Read 2 more answers

A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate

Romashka [77]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago