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Yuki888 [10]
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
Physics
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>

3 0
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
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