All categories

3 years ago

A race car starts from a rest to increases its speed to 23 m/s over a distance of 120 m. Calculate acceleration.

Physics

1 answer:

Orlov [11]3 years ago

6 0

Answer:

Acceleration=115m/s²

Explanation:

speed=distance/time

23=120/time

/120 on both sides

0.2s is time

acceleration= change in velocity/time taken

23/0.2=115

acceleration=115m/s²

You might be interested in

Another science question...

777dan777 [17]

In a neutral atom, the number of protons and electrons are always equal.

However, if there are more number of protons than electrons, we say the atom is now positively charged, (since protons are positively charged and the charges on electron cant cancel them out)

If there are more number of electrons than protons, the atom would be negatively charged.

If however there are different number of neutrons in an atom,but same proton number- we call them as isotopes. example 1H2 and 1H3 they have same protons but different neurtrons! hence they are isotopes of hydrogen

Hope this helpss

4 0

4 years ago

Read 2 more answers

How much heat energy is required to raise the temperature of 20 kilograms of water from 0°c to 35°c?

saveliy_v [14]

Energy required = Specific Heat × mass × change in temp.
Specific Heat (Cp) water = 4.186kJ/kgC 
Mass = 20kg 
Change in temperature = 35C - 0C = 35C 

Q = 4.186kJ/kgC × 20kg × 35C = 293.02kJ hope this helps! (:

6 0

3 years ago

Read 2 more answers

You have 50L of water, it froze. So find its new volume. Density of water is 1000kg/m^3 or 1kg/L, density of ice is 920kg/m^3 or

ad-work [718]

Answer:

D (density) = Mass / Volume

V (ice) = Mass / Density = 50000 g / .92 kg / L

V = 50 kg / .92 kg / L

1 Liter of water weighs 1 kilogram = 50 L weighs 50000 g

V = 54.3 L the mass does not change upon the change of phase (freezing)

4 0

3 years ago

A researcher is using a particle accelerator in an experiment studying isotopes. How can the researcher change one isotope into

Illusion [34]

The researcher can change one isotope into a different isotope of the same element by ADDING OR REMOVING NEUTRONS.
Isotopes refers to two or more forms of the same element, which have the same number of protons but different number of neutrons. The difference in the number of neutrons of a particular element is what brings about isotopes, thus, isotopes can be created by removing or adding neutrons to a particular element.

5 0

3 years ago

Read 2 more answers

Insert

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago