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Zinaida [17]
3 years ago
6

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

f 200 N. The box moves because of the two forces applied to it. Leon records the forces and direction of the forces acting on the box in his lab notebook. A 2 column table with 5 rows. Column 1 is labeled Force with entries Normal, Tension by Eduardo, Tension by Clara, Kinetic friction, Gravity. Column 2 is labeled Direction of Force Vector with entries Upward, Left, Right, Left, Downward. In the table, which force has the wrong direction? Tension by Eduardo Tension by Clara Kinetic friction Gravity
Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg =  weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

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A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin,
WINSTONCH [101]

Answer:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}

distance = 7.89 km

4 0
3 years ago
A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance
nlexa [21]

Answer:

22

Explanation:

6 0
3 years ago
‼️PLEASE ANSWER WILL MARK BRAINLIEST‼️​
Ahat [919]

Answer:

b 1000

if you multiply 2m/s by 500 you get 1000 it should look like this.

500 x 2m/s

6 0
2 years ago
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magni
d1i1m1o1n [39]

Answer:

a) 1.62*10^{-18}N

b) 2.84*10^{-19}N

c) 1.16*10^{-18}N

Explanation:

The net force is given by the expression:

\vec{F}=q(\vec{E}+\vec{v}\ X\ \vec{B})

By the right hand rule, we know that the direction of the magnetic force is

j X -i = k

\vec{v}X\vec{B}=vB\hat{k}

Hence, the direction of the magnetic force is the +z direction.

(a) E=4.19V/m k

F=(1.6*10^{-19}C)[4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=1.62*10^{-18}N

where we have used q=9.1*10^{-31}kg.

(b) E=-4.19V/m k

F=(1.6*10^{-19}C)[-4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=2.84*10^{-19}N

(c) E=4.19V/m i

In this case the net force will have two components:

F=(1.6*10^{-19}C)[4.19\frac{V}{m}\hat{i}+(2540\frac{m}{s})(2.35*10^{-3}T)\hat{k}]\\\\F=[6.704*10^{-19}\hat{i}+9.55*10^{-19}\hat{k}]N

and its magnitude will be:

F=\sqrt{(6.704*10^{-19})^2+(9.55*10^{-19})^2}=1.16*10^{-18}N

hope this helps!!

5 0
3 years ago
Please please help me on this
MaRussiya [10]

Answer:

Gravitational force has a long range but very small magnitude even there is a gravitational attraction between earth and sun.So, how massive is the body is also an important point to consider.

5 0
3 years ago
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