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2 years ago

7

Mr. Dunn drives 64.8km from work at a speed of 48km/h. Mrs. Dunn drives 81.2km from work

1 answer:

artcher [175]2 years ago

7 0

Answer:

i) Mr. Dunn arrives to home first.

ii) 3 min

Explanation:

i. To find who arrives first to home you calculate the time, by using the following formula:

$t=\frac{x}{v}$

x: distance

v: velocity

Mr. Dunn:

$t=\frac{64.8km}{48km/h}=1.35h$

Mrs. Dunn:

$t=\frac{81.2km}{58km/h}=1.4h$

Hence, Mr. Dunn arrives to home first.

ii. To calculate the difference in minutes, you convert hours to minutes:

$1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min$

the difference between the times is 3min

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Applying $\mu$ to the above equation and $\theta=10^{\circ}$

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2 years ago

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Gekata [30.6K]

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = $6.5*10^{4}\ m/s$

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity $[\theta]$ = 15 degree.

We are asked to calculate the magnetic force experienced by the charged particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force $\vec F=q(\ \vec V \times \vec B)$

i.e F = $=qVBsin\theta$

$=6.8*6.5*10^{4}*1.4*sin15$

$=61.88*10^{4}sin15$

$=61.88*10^4*0.2588\ N$

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