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3 years ago

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In order to get a tree stump out of the ground, chains are connected to two trucks. One truck pulls with a force of 600 N to the

north while another truck pulls with a force of 800 N to the west. What is the net force on the stump

1 answer:

Black_prince [1.1K]3 years ago

3 0

Answer:

The net force on the stump is 1000 N.

Explanation:

Given that,

Force 1 acting on the truck, $F_1=600\ N$ (due north)

Force 2 acting on the truck, $F_2=800\ N$ (due west)

We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :

$F=\sqrt{F_1^2+F_2^2}$

$F=\sqrt{600^2+800^2}$

F = 1000 N

So, the net force on the stump is 1000 N. Hence, this is the required solution.

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A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

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Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

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A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

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Answer:

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Explanation:

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Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

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Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

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A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro

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Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

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Put the value into the formula

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Using formula of flux

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Put the value into the formula

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Using formula of emf

$\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})$

Put the value into t5he formula

$\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})$

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$\epsilon=16.87\ mV$

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.016867}{2.5}$

$I=0.00675\ A$

$I=6.75\ mA$

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

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