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gavmur [86]
3 years ago
14

An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi

ll not move the object?
Physics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

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I need the evaluation
GalinKa [24]

Answer:

The maximum torque exerted by the person is, τ = 89.96 kg m² s⁻²

Explanation:

Given data,

The mass of the person riding a bike, m = 54 kg

The radius of rotation of the pedal, r = 17 cm

                                                       = 0.17 m

The torque exerted by the person is given by the formula,

                           τ = F r Sin θ

Where,

                          θ is the angle between F and R

The maximum torque exerted by the person is at θ = 90°

Therefore,

                            τ = F r

                                = m x g x r

                                = 54 x 9.8 x 0.17

                                = 89.96 kg m² s⁻²

Hence, the maximum torque exerted by the person is, τ = 89.96 kg m² s⁻²

3 0
3 years ago
What is a superpositional principle
natali 33 [55]
 the principle that in a series of stratified sedimentary rocks the lowest stratum is the oldest. 2. n. the displacement of any point due to the superposition <span>of wave systems is equal to the sum of the displacements of the individual waves at that point.</span>
7 0
3 years ago
Read 2 more answers
What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc
Murljashka [212]

Answer:

The Value is V_{AB} = 2.825V

Explanation:

The explanation is shown on the first  uploaded image

5 0
3 years ago
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
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