All categories

3 years ago

Hellppp meeee with this question

Physics

2 answers:

Reptile [31]3 years ago

7 0

Answer:

The acceleration of the rattlesnake, a = 126 m/s²

VLD [36.1K]3 years ago

5 0

Answer:

The acceleration of the rattlesnake, a = 126 m/s²

Explanation:

Given,

The initial velocity of the rattlesnake, u = 1.0 m/s

The final velocity of the rattlesnake, v = 20.0 m/s

The time duration to reach the final velocity, t = 0.15 s

The acceleration of a body defined as the rate of change of velocity by time, or simply the difference of final velocity and initial velocity by time.

The formula for acceleration is

a = (v-u)/t m/s²

Substituting the values in the above equation

a = (20 - 1.0 ) / 0.15

= 126 m/s²

Hence, the acceleration of the rattlesnake is, a = 126 m/s²

You might be interested in

1. In a chemical reaction, the forward rate is greater than the reverse rate. Which statement about the reaction is correct?

Alexeev081 [22]

Answer:

the reaction is reversible

5 0

3 years ago

What might say the first stage of photosynthesis powers the energy engine of the living world what do you think this sentence me

zzz [600]

I am having trouble on the same question.

8 0

3 years ago

Read 2 more answers

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

$\lambda = 3.68 *10^{-36} \ m$

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

A piece of aluminum with a mass of 3.05 g initially at a temperature of 10.8 °C is heated to a temperature of 20.

fgiga [73]

Answer:

25.3J

Explanation:

Given parameters:

Mass of aluminum = 3.05g

Initial temperature = 10.8 °C

Final temperature = 20 °C

Specific heat = 0.9J/g °C

Unknown:

Amount of heat needed for the temperature to change = ?

Solution:

To solve this problem, we use the expression:

H = m C Ф

H is the amount of heat

m is the mass

C is the specific heat capacity

Ф is the change in temperature

H = 3.05 x 0.901 x (20 - 10.8) = 25.3J

7 0

3 years ago

At approximately what wavelength of the continuous spectrum will the greatest (maximum) intensity occur when 60-kV electrons str

inna [77]

Answer: 20 pm=20*10^-12 m

Explanation: To solve this problem we have to use the relationship given by:

λmin=h*c/e*ΔV= 1240/60000 eV=20 pm

this expression is related with the bremsstrahlung radiation when a flux of energetic electrons are strongly stopped hitting to a catode. The electrons give their kinetic energy to the atoms of the catode.

6 0

3 years ago