What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2

Serhud [2]

Answer:

the potentail of kinetic and potential energy

Explanation:

first explain the concept of kinetic energy (what it is and what its used for) and give examples (cars, a basketball thrown across a hall, and airplane), and do the same with potential energy (the energy an object stores, example: a streched rubber band)

3 0

3 years ago

Read 2 more answers

A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, awa

igomit [66]

Answer:

At a distance of 1376.49 candle emits 0.2 watt power

Explanation:

Distance between Sun and earth $6.6\times 10^{16}m$

Sun emits a power of $P=3.8\times 10^{26}watt$

Power emitted by candle = 0.20 watt

We know that brightness is given by

$B=\frac{P}{4\pi d^2}$

So $\frac{3.8\times 10^{26}}{4\pi (6\times 10^{16})^2}=\frac{0.20}{4\pi d^2}$

$3.8\times 10^{26}d^2=7.2\times 10^{32}$

$d^2=1.89\times 10^6$

$d=1376.49m$

So at a distance of 1376.49 candle emits 0.2 watt power

3 0

3 years ago

A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from res

Bumek [7]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

$t_1 = \sqrt{\frac{2H}{g}}$

now the sound will come back to the observer in the time

$t_2 = \frac{H}{v}$

so we will have

$t_1 + t_2 = 2.42$

$\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42$

so we have

$\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42$

solve above equation for H

$H = 26.8 m$

Part b)

If sound reflection part is ignored then in that case

$H = \frac{1}{2}gt^2$

$H = \frac{1}{2}(9.81)(2.42^2)$

$H = 28.7 m$

so here percentage error in height calculation is given as

$percentage = \frac{28.7 - 26.8}{26.8} \times 100$

$percentage = 7.18$

5 0

3 years ago

Insert

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

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:

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

B)

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:

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

C)

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

A refrigerator

andrew11 [14]

From the definition of the first Law of thermodynamics we know that heat flows from the hottest object to the coldest. What has a refrigerator inside is a refrigerant that is traveling through some heat exchangers, driven by pumps and that allows the exchange of heat between a 'Hot' tank and then transport it to a certain place, eject it and return to extract more heat. In this way the refrigerant removes heat from a hot region to a cold region and heat enters said hot region. Since the heat extracted is greater than that of the medium, the heat from the inside flows outward.

The correct answer is B.

6 0

3 years ago