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3 years ago

8

Looking at this skier. If ht travels from point A to point B, then turns around and stops at point C, what is her total displace

ment?
A:180 m
B:320 m
C:100 m
D:40 m

2 answers:

trasher [3.6K]3 years ago

8 0

D. I just answered that question for someone else

never [62]3 years ago

8 0

D. 40 m is the correct answer

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A kite is hovering over the ground at the end of a straight 38-m line. the tension in the line has a magnitude of 17 n. wind blo

postnew [5]

Let the tension in the thread will be at angle theta

now by force balance in x and y directions we can say

$17 cos\theta = 18 cos60$

$17 cos\theta = 9$

$\theta = 58.03 degree$

now for the height of the kite we can use

$h = L sin\theta$

$h = 38 sin58.03$

$h = 32.23 m$

<em>so it will be at height 32.23 m from ground</em>

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3 years ago

Mike stands on a scale in an elevator. If the elevator is accelerating upwards with 4.9 m/s2, the scale reading is ____ times Mi

Bas_tet [7]

Answer:

F - M a force exerted by scales on student

M a = M (9.8 + 4.9) m/s^2 upwards chosen as positive

a = 1.5 g net acceleration of student due to force of scales

W =M g weight of student (actual weight)

Wapp = M 1.5 * g apparent weight (on scales) of student

7 0

2 years ago

Physics question, any help is appreciated :)

Gekata [30.6K]

Runner 2 sees Runner 1 passing him with a velocity of 17 m/s west.

8 0

3 years ago

Read 2 more answers

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago