What does 'change only one variable at a time' mean

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

Alexxandr [17]

Answer:

magnitude of A − B = 15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = $\sqrt{(-15)^{2}+(-5)^{2}}$

|A - B| = $\sqrt{250}$

|A - B| = 15.81 km

7 0

3 years ago

Please answer this question for me and explain why.

horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass $m_2$ and the 4kg mass $m_1$ . If the tension in the string is $T$ then for the mass $m_2$

(1). $T-m_2g=-m_2a$ <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass $m_1$

(2). $T-m_1g =m_1a$ <em> (the acceleration is upwards, hence the positive sign)</em>

Solving for $T$ in the 2nd equation we get:

$T =m_1a+m_1g$ ,

and putting this into the 1st equation we get:

$m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g$

$\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}$

Back to the question:

Using the formula for the acceleration we find

$a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g$

$a = \dfrac{g}{9},$

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Which one of the following types of electromagnetic wave travels through space the fastest?

Setler [38]

Answer:

As a result, light travels fastest in empty space, and travels slowest in solids. In glass, for example, light travels about 197,000 km/s.

Explanation:

4 0

2 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

oee [108]

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

3 0

3 years ago