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Ahat [919]
3 years ago
15

3) A charged particle is moving with velocity of V in a magnetic field of B, which one of the followings is correct: A) The dire

ction of force F on the charge is parallel to magnetic field B B) The direction of force F on the charge is parallel to velocity direction V C) The force is maximized when velocity direction and magnetic field are parallel D) The force F is perpendicular (normal) to both velocity V and magnetic field B E) The direction of force on positive charge or negative charge would be the same
Physics
1 answer:
Tcecarenko [31]3 years ago
7 0

Answer:

<em>D) The force F is perpendicular (normal) to both velocity V and magnetic field B.</em>

Explanation:

When a charged particle enters a magnetic field, it experiences a force which changes its direction of travel. The direction of motion of the charged particle, the magnetic field direction, and the direction of the force, are all perpendicular to one another. According to Lorentz right hand rule, hold the right hand parallel to the ground, with the palm facing up, and the thumb held out at right angle to the other fingers. If the direction of the other fingers represents the magnetic field line and direction, and the thumb represents the direction of motion of a positively charged particle, then, the palm will push up in the direction of the force. For a negatively charged particle, the force will push down in the direction of the back of the hand.

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Morgarella [4.7K]
Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
3 0
2 years ago
Displacement vectors of 4 km north, 2 km south, 5 km north, and 5 km south combine to a total displacement of
valentinak56 [21]

south = -(north)

Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)

Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)

Displacement = (4 - 2 + 5 - 5) km north

<u>Displacement = 2 km north </u>

6 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
What does the left y-axis show?
vivado [14]

Answer:

The correct option is;

a- sea surface temperature anomaly, in degrees Celsius

Explanation:

From the diagram related to the question we have two graphs super imposed of Sea surface temperature anomaly, in degrees Celsius and cholera incidence anomaly (%) both plotted against time in years.

On the left the y-axis represents the sea surface temperature anomaly while on the right, the y-axis represents the cholera incidence anomaly (%).

The display of the graph shows the sea surface temperature anomaly in blue.

6 0
2 years ago
The volume and the mass of substance are 15cm3 and 27 gm respectively find its density​
harina [27]

Answer:

d=1.8\ g/cm^3

Explanation:

Given that,

Mass, m = 27 grams

Volume of the substance, V = 15 cm³

We need to find the density of the substance. We know that, the density of an object is given by mass per unit volume. So,

d=\dfrac{m}{V}\\\\d=\dfrac{27}{15}\\\\d=1.8\ g/cm^3

So, the density of the substance is equal to 1.8\ g/cm^3.

8 0
3 years ago
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